Explanation:
Okay, so I need to find the measure of angle AFD in quadrilateral ABCD. The problem says that angles BAD and CDA are trisected, and there are some labels with x and y, as well as given angles of 110° and 100°. Let me try to visualize the diagram based on the Asymptote code provided.

First, let me sketch quadrilateral ABCD mentally. Points A, B, C, D are given with coordinates: A is at (0,0), B at (2,4), C at (7,4), and D at (7,-2). So, connecting A to B to C to D and back to A. The Asymptote code draws A to B to C to D to A. Then there are points E and F, with E near (4.5,1) and F at (5,3). The lines A-E-D and A-F-D are drawn, so E and F are points on the trisectors of angles BAD and CDA, respectively.

The labels mention angles labeled with x and y, and specific angles of 110° and 100°. The problem statement says angle BAD and angle CDA are trisected. Let me note that. So angle BAD is trisected into three equal parts, and angle CDA is also trisected into three equal parts. The Asymptote code labels some parts with x and y. Let me try to figure out how they are placed.

Looking at the Asymptote code:

Label "x" at (1, 1.5) and (2, 1), (2, -1). The label "y" at (5.5, 0.5), (6.5, 0), (5.5, -1.5). Then labels "110°" near point B and "100°" near point C. Wait, the Asymptote code says:

label("$110^{\circ}$",(2.5,3.5)); label("$100^{\circ}$",(6.5-.2,3.5));

So, those labels seem to be near the angles at B and C. Wait, the coordinates (2.5,3.5) is near point B which is at (2,4), so maybe angle at B is 110°, and angle at C is 100°? Wait, but angle at C is between points B, C, D. Wait, but the label for 100° is at (6.5 - 0.2, 3.5) which is (6.3, 3.5). Since point C is at (7,4), so that's near point C. So maybe angle at B is 110°, angle at C is 100°? But the problem didn't specify those angles, but they are given in the Asymptote code. Wait, but maybe the original problem has those angles, but the Asymptote code is just an illustration? Wait, the problem says "angle BAD and angle CDA are trisected as shown." So perhaps the diagram is drawn with specific angles, and the labels x, y, and 110°, 100° are part of the diagram. Maybe angle BAD is trisected into three angles, each labeled x, and angle CDA is trisected into three angles labeled y? Wait, looking at the Asymptote code:

label("$x$", (1, 1.5), S); label("$x$", (2, 1), S+W); label("$x$", (2, -1), N+N+N+W);

So there are three x's. Let me see their coordinates. The first x is at (1, 1.5), which is near the middle of AB? Because A is at (0,0), B is at (2,4). So (1,1.5) is somewhere between A and B? Wait, but angle BAD is at point A between BA and DA. Wait, angle BAD is the angle at A between AB and AD. Similarly, angle CDA is the angle at D between DC and DA.

So, if angles BAD and CDA are trisected, then angle BAD is divided into three equal angles, and angle CDA is divided into three equal angles. The Asymptote code draws lines from A to E and from A to F. Wait, but E is connected to D, so maybe E is on the trisector of angle BAD? Similarly, F is connected to D, so F is on the trisector of angle CDA? Or maybe E is on the trisectors of angle BAD and F is on trisectors of angle CDA?

Wait, the Asymptote code says:

draw(A--E--D); draw(A--F--D);

So E is connected to A and D, so E is a point inside the quadrilateral connected to A and D. Similarly, F is connected to A and D. Wait, but A-F-D would be a line from A to F to D? But then if F is on the trisector of angle CDA, maybe?

Wait, maybe E is one of the trisectors of angle BAD and F is one of the trisectors of angle CDA? Let me think.

First, let me figure out the angles at A and D. The problem mentions trisection, so maybe angle BAD is trisected into three angles, each of measure x, and angle CDA is trisected into three angles, each of measure y. Then the labels x and y might be those trisected angles? But the Asymptote code labels three x's and three y's. Let me try to see.

But maybe looking at coordinates can help. Let me compute coordinates of points to figure out angles.

Coordinates:

A = (0,0)

B = (2,4)

C = (7,4)

D = (7,-2)

So AB is from (0,0) to (2,4). AD is from (0,0) to (7,-2). So angle BAD is the angle between AB and AD.

Similarly, angle CDA is the angle at D between DC and DA. DC is from D to C, which is from (7,-2) to (7,4), so that's straight up along x=7. DA is from D to A, which is from (7,-2) to (0,0), so that's a line going to the left and up.

So angle CDA is between DC (upwards vertical) and DA (line from D to A). Let me try to compute angles BAD and CDA first.

But maybe it's better to find angle BAD first. Let me calculate vectors.

For angle BAD at point A between AB and AD.

Vector AB is (2,4) - (0,0) = (2,4)

Vector AD is (7,-2) - (0,0) = (7,-2)

So angle between AB and AD can be calculated using the dot product.

Dot product of AB and AD is 2*7 + 4*(-2) = 14 - 8 = 6

The magnitude of AB is sqrt(2^2 + 4^2) = sqrt(4 + 16) = sqrt(20) = 2*sqrt(5)

The magnitude of AD is sqrt(7^2 + (-2)^2) = sqrt(49 + 4) = sqrt(53)

So cos(theta) = 6 / (2*sqrt(5)*sqrt(53)) = 6 / (2*sqrt(265)) = 3 / sqrt(265)

But maybe that's not necessary. Wait, maybe we can find angle BAD?

Alternatively, perhaps using coordinates to find the slopes and compute the angle.

Slope of AB: from A(0,0) to B(2,4): slope is (4-0)/(2-0) = 2

Slope of AD: from A(0,0) to D(7,-2): slope is (-2 - 0)/(7 - 0) = -2/7

So angle between two lines with slopes m1 and m2 is given by:

tan(theta) = |(m2 - m1)/(1 + m1*m2)|

So tan(theta) = |(-2/7 - 2)/(1 + (2)*(-2/7))| = |(-2/7 - 14/7)/(1 - 4/7)| = |(-16/7)/(3/7)| = | -16/3 | = 16/3

So theta is arctangent of 16/3. Let me compute that. Well, arctan(16/3) is some angle. Let me note that 16/3 ≈ 5.333, so arctan(5.333) is roughly 79 degrees? Wait, tan(80 degrees) is approx 5.671, so arctan(16/3) ≈ 79 degrees? Wait, tan(79°) ≈ 5.145, so 16/3 is about 5.333, so angle is a bit more. Maybe around 79.5 degrees? Let me check:

tan(79°) ≈ 5.145

tan(79.5°): let's compute:

Using calculator, tan(79.5 degrees) ≈ tan(79 + 0.5) ≈ ?

But maybe I can calculate it. Alternatively, maybe I can find exact value? Not sure. Maybe it's not necessary right now.

Similarly, angle CDA. Let me compute angle at D between DC and DA.

Vector DC is from D to C: (7,4) - (7,-2) = (0,6). So vector DC is straight up along positive y-axis.

Vector DA is from D to A: (0,0) - (7,-2) = (-7, 2). So vector DA is (-7, 2). So angle between DC (0,6) and DA (-7, 2).

So, angle between vector DC (0,6) and DA (-7,2). Let me calculate that angle.

First, vector DC is (0,6), direction is straight up. Vector DA is (-7,2). The angle between DC and DA can be calculated using dot product.

Dot product: (0)(-7) + (6)(2) = 12

Magnitude of DC is 6.

Magnitude of DA is sqrt( (-7)^2 + 2^2 ) = sqrt(49 +4) = sqrt(53)

So cos(theta) = 12 / (6 * sqrt(53)) ) = 2 / sqrt(53)

Therefore theta = arccos(2 / sqrt(53)) ≈ arccos(0.277). Let me compute that. sqrt(53) ≈ 7.28, so 2 / 7.28 ≈ 0.275. arccos(0.275) ≈ 73.7 degrees. Hmm, so angle CDA is approximately 73.7 degrees. But maybe exact value? Let me think.

Wait, but maybe the problem does not require us to compute these angles directly. Instead, given that angles BAD and CDA are trisected, and there are some labels of x and y, as well as given angles of 110° and 100°, which are probably angles at B and C? The Asymptote code labels "110°" at (2.5,3.5) and "100°" at (6.5 - 0.2, 3.5). Given that point B is at (2,4), so (2.5,3.5) is near B, so that's likely angle at B. Similarly, point C is at (7,4), so label "100°" is near C, so angle at C is 100°, and angle at B is 110°? Wait, but in quadrilaterals, the sum of angles is 360°, so if we can find other angles, we can check.

But let me check the given data. So quadrilateral ABCD. Angles at B and C are given as 110° and 100°, and angles at A and D are being trisected. If I can find angles at A and D, then maybe I can find the trisection angles x and y, then find angle AFD.

Alternatively, since angle BAD is trisected into three x's and angle CDA is trisected into three y's. So angle BAD = 3x, angle CDA = 3y. Then, in quadrilateral ABCD, the sum of internal angles is 360°, so angle A + angle B + angle C + angle D = 360°. If we can write that equation with angles A = 3x, angle D = 3y, angle B = 110°, angle C = 100°, then 3x + 110 + 100 + 3y = 360. Therefore, 3x + 3y = 360 - 210 = 150. So x + y = 50°. So, x + y = 50°. That's one equation.

But we need more equations to find x and y. However, the Asymptote code labels x's and y's. Let me see. The diagram labels three x's and three y's. Let me think about how the trisection occurs. So angle BAD is trisected into three angles, each x. So if angle BAD is at vertex A, between AB and AD, then the trisectors would divide that angle into three equal parts. Similarly, angle CDA is at vertex D, between DC and DA, trisected into three angles each y. So if I can find relations based on the diagram, maybe involving triangle AFD or other triangles.

Alternatively, perhaps using the fact that lines AE and AF are trisectors. Wait, in the Asymptote code, they draw A--E--D and A--F--D. So E is connected to A and D, and F is connected to A and D. So AE and AF are trisectors of angle BAD and angle CDA?

Wait, angle BAD is at A, so if AE is one of the trisectors, then angle between AB and AE is x, angle between AE and AF is x, and angle between AF and AD is x. Similarly, for angle CDA at D, trisected into three angles y, so the trisectors would be from D, but the diagram shows lines from A to F to D. Wait, unless F is a point on the trisector of angle CDA, meaning that DF is a trisector?

But the Asymptote code draws A--F--D. So point F is connected to A and D. So maybe F is on the trisector of angle CDA? Wait, angle CDA is at D, so the trisectors would be from D. However, the diagram shows a line from A to F to D. Therefore, maybe AF is a trisector of angle BAD, and FD is a trisector of angle CDA? But that would mean FD is a trisector of angle CDA. But angle CDA is at D, so trisectors would be lines from D. Therefore, maybe FD is a trisector of angle CDA, meaning that DF divides angle CDA into y and 2y? Or something like that?

Wait, this is getting confusing. Let me try again. Let me think about trisectors.

Angle BAD is at A, between BA and DA. If it is trisected, then two trisectors are drawn, dividing angle BAD into three equal angles. Similarly, angle CDA is at D, between CD and DA, trisected into three equal angles.

In the diagram, they have lines from A to E and A to F. So maybe E is one trisector of angle BAD, and F is one trisector of angle CDA? Wait, but if F is a trisector of angle CDA, which is at D, then the line DF would be the trisector. But the diagram shows A-F-D. Therefore, AF is a line from A to F, which is connected to D. Therefore, AF is not a trisector of angle CDA, but perhaps of angle BAD?

Wait, angle BAD is at A, so trisectors would be lines from A. Therefore, AE and AF are trisectors of angle BAD? Then there would be two trisectors, dividing BAD into three parts. Similarly, angle CDA is at D, so trisectors would be lines from D. But in the diagram, the lines from D would be... the diagram shows A-F-D, so FD is connected to D, which could be a trisector. So FD is a trisector of angle CDA. So F is a point such that DF trisects angle CDA. So then AF connects to F, which is somewhere inside the quadrilateral.

Alternatively, maybe E is a trisector of angle BAD and F is a trisector of angle CDA. So AE is a trisector of angle BAD, and DF is a trisector of angle CDA. But the diagram has A-E-D and A-F-D. So E is between A and D, connected to A and D? Wait, if E is on the trisector of angle BAD, then it's a line from A going into the quadrilateral, but E is connected to D? So E is connected to A and D, so it's a line from E to D. Therefore, E lies somewhere such that AE is a trisector of angle BAD, and ED is a line from E to D. Similarly for F.

This is getting too vague. Maybe I need to use the given angles of 110° and 100°, and the x and y labels. Let me think again about the earlier equation: x + y = 50°, from angle sum of quadrilateral. If I can find another equation relating x and y, I can solve for x and y, and then use that to find angle AFD.

Alternatively, maybe triangle AFD is involved. Let me think about triangle AFD. To find angle AFD, we need to know something about triangle AFD. If I can find the measures of other angles in triangle AFD or some relations, maybe using the trisection angles.

But how?

Alternatively, using the given labels of x and y. The Asymptote code labels three x's and three y's. Let me try to figure out where they are placed. The first label for x is at (1,1.5), which is near AB. Second x at (2,1) S+W, which is near the middle of AB and AD? Third x at (2,-1), N+N+N+W, which is near AD? Similarly, y's are at (5.5+.3, .5-.3), which is near DC and DA. Maybe these labels correspond to the trisection angles?

Let me think. For angle BAD, which is at A, if it's trisected into three angles of x each, then maybe the labels x are on those trisected angles. Similarly for angle CDA trisected into three angles of y each.

But in the diagram, the Asymptote code draws A to E to D and A to F to D. So E is connected to D, which is part of angle CDA? Or is E connected to D as part of angle BAD?

Wait, if angle BAD is trisected, then the trisectors would be two lines from A dividing angle BAD into three equal parts. Similarly, angle CDA is trisected, so two lines from D dividing angle CDA into three equal parts. However, in the diagram, there are two lines from A: AE and AF, and since E and F are connected to D, that suggests that E and F are points on those trisectors. Wait, but if angle BAD is trisected, then the trisectors would be two lines from A, but in the diagram, they have two lines from A: AE and AF. If AE and AF are the trisectors of angle BAD, then they divide angle BAD into three parts. Similarly, if there is another line from D for angle CDA.

But the diagram shows A-F-D. If F is connected to D, then maybe DF is one of the trisectors of angle CDA. Therefore, DF divides angle CDA into two parts, one being y and the other being 2y or something? Wait, but if angle CDA is trisected, then there should be two trisectors dividing it into three equal angles. So if DF is one of the trisectors, then the angle between DC and DF is y, and between DF and DA is 2y? Or perhaps between DC and DF is y, DF to DA is y, and then another trisector? Wait, no, if it's trisected, there are two trisectors, dividing the angle into three equal parts. So angle between DC and first trisector is y, between the two trisectors is y, and between the second trisector and DA is y. Therefore, the total angle CDA is 3y.

But in the diagram, the line from D is DF. If DF is one of the trisectors, then maybe FD is the first trisector, dividing angle CDA into y and 2y, but without more information, it's hard to tell.

Alternatively, since the diagram labels x and y in certain locations, perhaps those x's are the trisected angles at A, and y's at D. Let me think. The three x's are at (1,1.5), (2,1), and (2,-1). The (1,1.5) is near AB, so maybe that is one of the trisection angles between AB and AE? Then the next x at (2,1) is near AB and AD? Wait, this is too vague. Maybe I need to consider that the angles labeled x are adjacent to AB and AD? Let me try to imagine the diagram again.

Since I can't see the image, but based on Asymptote code, maybe the Asymptote code draws A to E to D, so E is inside the quadrilateral, connected to A and D. Similarly, F is connected to A and D. The labels x are on three different positions. Let me try to think again about the angles.

Given that angle BAD is trisected into three x's, so if angle BAD is 3x, and angle CDA is 3y, and from earlier, x + y = 50°. So if I can find another relation between x and y, I can solve for them. How?

Perhaps using triangle AFD. If I can find angles at A and D related to triangle AFD, then I can find angle at F.

But angle at A for triangle AFD is angle FAD, which is part of angle BAD. Similarly, angle at D for triangle AFD is angle FDA, which is part of angle CDA. Depending on how AE and AF are placed, those angles could be x or multiples thereof.

Wait, assuming that AF is a trisector of angle BAD, then angle FAD would be x or 2x? Similarly, if FD is a trisector of angle CDA, then angle FDA would be y or 2y?

But without knowing exactly which trisectors AE and AF are, it's hard. However, the Asymptote code labels three x's and three y's. If there are three x's, that suggests that angle BAD is divided into three equal parts, each x. Similarly for angle CDA with three y's. So if angle BAD = 3x, angle CDA = 3y. Then, from quadrilateral angle sum, 3x + 3y + 110 + 100 = 360 => 3x + 3y = 150 => x + y = 50° as before.

But how can I find another equation?

Maybe looking at triangle AFD. In triangle AFD, the sum of angles is 180°, so angle FAD + angle FDA + angle AFD = 180°. If I can find angle FAD and angle FDA in terms of x and y, then I can find angle AFD.

If AF is a trisector of angle BAD, then angle FAD could be 2x or x, depending on which trisector it is. Similarly, if FD is a trisector of angle CDA, then angle FDA could be 2y or y.

But the diagram has two lines from A: AE and AF. If AE and AF are the two trisectors of angle BAD, then they divide angle BAD into three x's. So if AF is the first trisector, then angle between AB and AF is x, and between AF and AE is x, and between AE and AD is x. But then the Asymptote code draws A--E--D and A--F--D. So if AF is a trisector, then maybe angle between AF and AD is x? Wait, if the trisection is from AB to AD, then the first trisector would divide angle BAD into x and 2x. The second trisector divides into 2x and x. So depending on which trisector AF is, angle FAD could be 2x or x. Similarly for FD.

But how can I determine?

Alternatively, looking at the labels in the diagram. The Asymptote code labels three x's. If angle BAD is trisected into three angles, each adjacent to AB, then perhaps the angles labeled x are between AB and AE, AE and AF, and AF and AD? But the diagram has three x's. So if AE and AF are the two trisectors, then there are three angles: AB to AE (x), AE to AF (x), AF to AD (x). Similarly for angle CDA.

But in the Asymptote code, the labels for x are at (1,1.5), (2,1), (2,-1). Let me try to think where those could be. The point (1,1.5) is between A(0,0) and B(2,4), and closer to A. The point (2,1) is between A and B? Or is it between AB and AD? Wait, maybe (2,1) is near the middle of AB and AD? Wait, AB goes from (0,0) to (2,4). AD goes from (0,0) to (7,-2). The point (2,1) is somewhere in the middle of the quadrilateral. Maybe it's the intersection of some lines?

Alternatively, maybe the three x's are the three angles formed by the trisection of angle BAD. So, if angle BAD is trisected, there are two lines from A dividing it into three equal parts. Then, each of those angles is labeled x, and those labels are at positions corresponding to those angles.

Similarly, three y's for angle CDA.

But without seeing the diagram, it's challenging, but maybe I can think of the trisection angles.

Alternatively, maybe using the given angles at B and C. The problem mentions "as shown," and in the Asymptote code, the angles at B is 110°, angle at C is 100°, so maybe we can compute angles at A and D?

Earlier, I calculated angle BAD ≈ arctan(16/3) ≈ 79 degrees? Wait, but if angle BAD is trisected, and angle CDA is trisected, but we have from quadrilateral sum that 3x + 3y = 150, so x + y = 50. If I could compute angles at A and D from coordinates, then I can find x and y.

Let me try that. Earlier, I tried computing angle BAD:

Vectors AB = (2,4), AD = (7,-2). The angle between them was calculated using tan(theta) = 16/3 ≈ 5.333, which gave theta ≈ arctan(5.333). Let me compute this angle more accurately.

Let me recall that tan(theta) = 16/3. So theta is arctan(16/3). Let me convert this to degrees.

Using calculator input:

arctan(16/3) ≈ arctan(5.333) ≈ 79.3 degrees.

Similarly, angle CDA was calculated as arccos(2 / sqrt(53)) ≈ arccos(0.277) ≈ 73.7 degrees.

So adding up, angle A ≈ 79.3°, angle B = 110°, angle C = 100°, angle D ≈ 73.7°. Let me check if they add up to 360:

79.3 + 110 + 100 + 73.7 ≈ 79.3 + 110 = 189.3; 189.3 + 100 = 289.3; 289.3 + 73.7 = 363°. Hmm, off by 3°, probably due to approximation errors in calculations.

But let me try to compute angle BAD and CDA more accurately.

First, angle BAD:

Vectors AB = (2,4) and AD = (7,-2).

The formula for the angle between two vectors is:

cos(theta) = (AB · AD) / (|AB| |AD|)

AB · AD = 2*7 + 4*(-2) = 14 -8 = 6

|AB| = sqrt(2^2 + 4^2) = sqrt(20) = 2*sqrt(5)

|AD| = sqrt(7^2 + (-2)^2) = sqrt(49 + 4) = sqrt(53)

Thus,

cos(theta) = 6 / (2√5 * √53) = 6 / (2√(265)) = 3 / √265 ≈ 3 / 16.2788 ≈ 0.1844

So theta ≈ arccos(0.1844) ≈ 80 degrees? Let me check:

cos(80°) ≈ 0.1736

cos(79°) ≈ 0.1908

So 0.1844 is between 79° and 80°, closer to 79.5°. Let me compute:

Let me compute cos(79.5°):

Using calculator, cos(79.5°) ≈ cos(79°30') ≈ 0.1822

Hmm, so 0.1844 is slightly higher. Let me use linear approximation.

Between 79° (cos ≈ 0.1908) and 80° (≈ 0.1736). Wait, the value 0.1844 is between 0.1736 and 0.1908? Wait no, 0.1844 is between 0.1736 and 0.1908? Wait, 0.1736 is 80°, 0.1908 is 79°. So cos(theta) = 0.1844 is between 79° and 80°, closer to 80°? Wait, no:

Wait, as theta increases from 0° to 90°, cos(theta) decreases from 1 to 0. So if cos(theta) = 0.1844, which is less than 0.1908 (cos79°), then theta is greater than 79°, since cos(theta) decreases. Therefore, theta is between 79° and 80°, with cos(theta) = 0.1844. Let me find it more precisely.

Let me denote theta = 79° + d, where d is in degrees between 0 and 1. Then cos(theta) = cos(79° + d) = cos(79°)cos(d) - sin(79°)sin(d). But maybe using linear approximation is better.

Let me approximate. Let me take theta ≈ 79.5°, as before, and compute cos(79.5°):

Actually, let me use calculator-like computation:

We can use the formula:

cos(a + b) ≈ cos(a) - b*sin(a) if b is small in radians.

Let me convert 79° to radians: 79° * π/180 ≈ 1.3788 radians.

Let me take a = 79°, and let me compute cos(a) ≈ 0.1908.

Let me let b be the difference in radians so that cos(a + b) = 0.1844.

So:

cos(a + b) ≈ cos(a) - b*sin(a) = 0.1844.

We have:

0.1908 - b*sin(79°) = 0.1844.

sin(79°) ≈ 0.9816.

Therefore:

0.1908 - 0.9816*b = 0.1844

=> 0.1908 - 0.1844 = 0.9816*b

=> 0.0064 = 0.9816*b

=> b ≈ 0.0064 / 0.9816 ≈ 0.0065 radians.

Convert 0.0065 radians to degrees: 0.0065 * (180/π) ≈ 0.0065 * 57.3 ≈ 0.372 degrees.

Therefore theta ≈ 79° + 0.372° ≈ 79.37°, so approximately 79.4°.

Similarly, angle at D: angle CDA. Earlier, we found cos(theta) = 2 / sqrt(53) ≈ 2 / 7.28 ≈ 0.2747. Therefore theta ≈ arccos(0.2747). Let me compute this.

cos(74°) ≈ 0.2756, which is very close to 0.2747. Therefore, angle CDA ≈ 74°, since cos(74°) ≈ 0.2756, and our value is slightly less, so angle is slightly more than 74°, say 74.1° or so.

Therefore, angles at A ≈ 79.4°, angle at D ≈ 74°, angle at B is 110°, angle at C is 100°, sum ≈ 79.4 + 110 + 100 + 74 ≈ 363.4°, which is off by about 3.4°, likely due to approximation errors in angle CDA. But if we take angle BAD ≈ 79.4°, angle CDA ≈ 74°, then 3x ≈ 79.4° => x ≈ 26.5°, 3y ≈ 74° => y ≈ 24.7°, so x + y ≈ 51.2°, which is close to our earlier result of x + y = 50°, considering rounding errors. So approximately x + y = 50°, which seems consistent.

But how does this help me? If I need to find angle AFD. If I can find angles FAD and FDA in triangle AFD, then angle AFD = 180° - angle FAD - angle FDA.

Assuming that AF is a trisector of angle BAD, then angle FAD could be x or 2x depending on which trisector it is. Similarly, if FD is a trisector of angle CDA, then angle FDA could be y or 2y.

But how do I know which trisectors AF and FD are?

Looking back at the Asymptote code: there is a label for x at (1,1.5) which is near AB, and another x at (2,1) which is closer to AD? Maybe those are the trisection angles. Similarly for y. Since there are three x's, it's likely that angle BAD is divided into three equal parts, each labeled x, and similarly for angle CDA.

Assuming that AF is the first trisector from A, so angle between AB and AF is x, then angle FAD would be 2x, but depending on how many trisection lines there are. Wait, angle BAD is trisected into three equal parts, so there are two trisectors. If AF is one trisector, then AF divides angle BAD into x and 2x. Similarly, the other trisector divides it into 2x and x. So if AF is the first trisector, then angle between AB and AF is x, and between AF and AD is 2x. Similarly, if FD is a trisector of angle CDA, then angle between DC and DF is y, and between DF and DA is 2y. Therefore, angle FDA would be 2y.

Therefore, in triangle AFD, angle FAD = 2x, angle FDA = 2y, and angle AFD = 180 - 2x - 2y = 180 - 2(x + y). Since we know x + y = 50°, then angle AFD = 180 - 2*50 = 180 - 100 = 80°. Wait, so angle AFD is 80°? But let me verify if that makes sense.

Alternatively, if angle FAD is x and angle FDA is y, then angle AFD would be 180 - x - y = 180 - 50 = 130°, which doesn't seem right. But why?

But this depends on which trisectors AF and FD are. If AF is dividing angle BAD into x and 2x, and FD divides angle CDA into y and 2y, then angle FAD is 2x and angle FDA is 2y. If that's the case, then angle AFD is 180 - 2x - 2y = 180 - 2*50 = 80°, which would be the answer. But is that correct?

Alternatively, if AF is the second trisector, then angle FAD = x and angle FDA = y, then angle AFD would be 180 - x - y = 130°, which is different. So how do I know which one it is?

Looking back at the Asymptote code, the labels for x's are at positions that may indicate where the trisectors are. For example, label x at (1,1.5): this is near the area between AB and AD? Wait, (1,1.5) is on the line from A(0,0) to B(2,4), closer to A? Wait, the coordinates (1,1.5) is halfway between A(0,0) and (2, 3), which is not exactly on AB. Wait, AB is from (0,0) to (2,4), so parametric equations. Let me check if (1,1.5) is on AB. If we go from A(0,0) to B(2,4), then at t=0.5, the point would be (1, 2). But (1, 1.5) is lower than that. Therefore, (1,1.5) is not on AB. Similarly, (2,1): is that on AD? AD is from (0,0) to (7,-2). Let me check if (2,1) is on AD. The parametric equation for AD is x = 7t, y = -2t for t from 0 to 1. So at t=2/7, x=2, y=-4/7 ≈ -0.57. But (2,1) is not on AD. Similarly, (2,-1): let's check if that's on AD. For y=-1, solve -2t = -1 => t = 0.5, then x=7*0.5=3.5. So (3.5, -1) is on AD. But (2,-1) is not. So these x's are not on the sides, but rather inside the quadrilateral. So perhaps they are labels for angles?

If angle BAD is trisected into three angles, maybe the trisectors are inside the quadrilateral, and the labels x are indicating those angles. So if I imagine angle BAD being trisected, then there are two lines from A dividing it into three equal parts. The diagram shows lines AE and AF. If those are the trisectors, then each trisection angle is x. Suppose that angle between AB and AE is x, angle between AE and AF is x, angle between AF and AD is x. Then, angle FAD is 2x. Similarly, if angle CDA is trisected by FD into three angles y, then the angle between DC and DF is y, between DF and FD... Wait, DF is a trisector. If FD is the first trisector, then angle between DC and FD is y, and between FD and DA is 2y. Therefore, angle FDA is 2y. Thus, angle AFD = 180 - 2x - 2y = 180 - 2(x + y) = 180 - 2*50 = 80°. Therefore, angle AFD is 80 degrees.

But how do I confirm this? Because if the labels x are the trisection angles, then each trisection angle is x, so angle between AB and AE is x, angle between AE and AF is x, angle between AF and AD is x. Similarly, for angle CDA, the trisection angles are y. Therefore, the diagram would have three angles labeled x for angle BAD and three labeled y for angle CDA.

Therefore, assuming that, then in triangle AFD, angle at A is angle FAD = 2x (since from AB to AF is x, then AF to AD is another x, total 2x), and angle at D is angle FDA = 2y (from DC to DF is y, then DF to DA is another y, total 2y). Therefore, angle AFD = 180 - 2x - 2y = 80°, as before.

Alternatively, if angle FAD was x and angle FDA was y, then angle AFD = 180 - x - y = 130°, but that would require that AF is one trisection angle and FD is one trisection angle, which might not be the case.

But given that there are three x's, it seems more likely that angle BAD is divided into three equal parts, each labeled x, and similarly for angle CDA with three y's. Therefore, the angle FAD would be 2x and angle FDA would be 2y, leading to angle AFD = 80°. Let me check if this answer makes sense with the diagram.

Alternatively, maybe there's another approach using the given angles at B and C. Since we know angles at B and C, maybe we can find the other angles.

Wait, earlier approximate calculations suggested angle BAD ≈ 79.4°, angle CDA ≈ 74°, so if angle BAD is 3x ≈ 79.4°, then x ≈ 26.5°, and angle CDA is 3y ≈ 74°, y ≈ 24.7°, so angle AFD = 180 - 2x - 2y ≈ 180 - 2*(26.5 + 24.7) = 180 - 2*(51.2) = 180 - 102.4 = 77.6°, which is close to 80°, considering the approximation. So that seems consistent.

Alternatively, if we use exact values, since we have x + y = 50°, then angle AFD = 180 - 2(x + y) = 180 - 100 = 80°, which is exact. Therefore, regardless of the exact values of x and y, as long as x + y = 50°, angle AFD would be 80°. Therefore, maybe the answer is 80 degrees.

But wait, I need to make sure that angle FAD is 2x and angle FDA is 2y. Is there a way to confirm that?

Let me think about the diagram again. Since AF is a trisector of angle BAD, which is 3x, then if there are two trisectors, dividing it into three parts. If F is closer to A or D?

Looking at the coordinates of point F: in the Asymptote code, F is at (5, 3). Let me see where that is. Point A is at (0,0), B at (2,4), C at (7,4), D at (7,-2). Point F is at (5,3), which is inside the quadrilateral, closer to the middle. Let me imagine line AF: from A(0,0) to F(5,3). Let me calculate the angle that AF makes with AB and AD.

Vector AF is (5,3). Vector AB is (2,4). Vector AD is (7,-2). Let me compute the angle between AF and AB, and between AF and AD.

First, angle between AF and AB:

Vectors AF = (5,3), AB = (2,4).

Dot product: 5*2 + 3*4 = 10 + 12 = 22

|AF| = sqrt(25 + 9) = sqrt(34) ≈ 5.830

|AB| = sqrt(4 + 16) = sqrt(20) ≈ 4.472

cos(theta) = 22 / (sqrt(34)*sqrt(20)) ≈ 22 / (5.830 * 4.472) ≈ 22 / 26.08 ≈ 0.843

Therefore theta ≈ arccos(0.843) ≈ 32.5°

Similarly, angle between AF and AD:

Vectors AF = (5,3), AD = (7,-2).

Dot product: 5*7 + 3*(-2) = 35 - 6 = 29

|AF| = sqrt(34) ≈ 5.830

|AD| = sqrt(49 + 4) = sqrt(53) ≈ 7.28

cos(theta) = 29 / (5.830 * 7.28) ≈ 29 / 42.43 ≈ 0.683

Therefore theta ≈ arccos(0.683) ≈ 47°

So angle between AF and AB is ≈ 32.5°, angle between AF and AD is ≈ 47°, and the total angle BAD is ≈ 79.5°, which matches. So 32.5 + 47 ≈ 79.5°, which is consistent. Therefore, if angle BAD is trisected into three equal parts, then each trisection angle would be 79.5° / 3 ≈ 26.5°, which matches our previous calculation.

So if angle between AF and AB is 32.5°, which is more than 26.5°, that suggests that AF is not the first trisector (which would be x ≈ 26.5° from AB), but rather the second trisector. Because if the first trisector divides angle BAD into x ≈ 26.5° from AB, then the second trisector would be another x ≈ 26.5° from the first trisector, totaling 53°, leaving the remaining angle to AD as 79.5 - 53 = 26.5°. Wait, but angle between AF and AB is 32.5°, which is larger than x, so that would mean that AF is located between the first trisector and the second trisector? Wait, but there are only two trisectors dividing angle BAD into three equal parts. So if angle between AB and first trisector is x, between first and second trisector is x, between second trisector and AD is x.

Therefore, if AF is located such that the angle between AB and AF is greater than x, then AF must be the second trisector. But according to the calculation, angle between AB and AF is ≈32.5°, which would be 2x, since if x ≈26.5°, then 2x ≈53°, but that's not matching. Wait, wait. Wait, if angle BAD is 3x ≈ 79.5°, then each x ≈26.5°, so from AB, the first trisector is at 26.5°, then the second trisector is at another 26.5°, totaling 53°, then the remaining to AD is 26.5°. Therefore, if AF is creating an angle of 32.5° with AB, which is between the first and second trisectors?

Wait, but according to calculation, angle between AF and AB is ≈32.5°, which would be between the first trisector (26.5°) and second trisector (53°). That suggests that AF is not one of the trisectors, but a different line. But according to the Asymptote code, AF is drawn as a trisector. There's a contradiction here. Maybe my assumption that AF is a trisector is wrong?

Alternatively, maybe angle between AF and AD is 47°, which is roughly 2x (since 2*26.5=53). Close to 47? There's discrepancy here.

Alternatively, maybe my calculation is wrong. Let me recheck.

First, angle between AF and AB: vectors AF=(5,3) and AB=(2,4). The angle between them:

cos(theta) = (AF · AB) / (|AF| |AB|) = (10 + 12)/ (sqrt(34)*sqrt(20)) = 22 / (sqrt(680)) ≈ 22 / 26.07 ≈ 0.844. arccos(0.844) ≈ 32.5°, that's correct. Similarly, angle between AF and AD:

AF · AD = 35 - 6 = 29, |AF|=sqrt(34), |AD|=sqrt(53). So cos(theta)=29 / (sqrt(34*53)) = 29 / sqrt(1802) ≈ 29 / 42.45 ≈ 0.683. arccos(0.683) ≈ 47°, correct.

Therefore, angle between AF and AB is ≈32.5°, angle between AF and AD is ≈47°, and angle BAD is ≈79.5°, which checks out because 32.5 + 47 ≈ 79.5°.

If angle BAD is trisected into three equal parts x, each ≈26.5°, then the first trisector is at ≈26.5° from AB, and second trisector at ≈53° from AB. But AF is at ≈32.5° from AB, which is between the two trisectors. Therefore, AF is not a trisector, which contradicts the problem statement that angles BAD and CDA are trisected as shown. Therefore, my assumption that AF is a trisector must be wrong. Hence, my previous reasoning is flawed.

Alternative approach: Maybe AE and AF are the trisectors. Let me check point E. In the Asymptote code, E is at (4.5 - 0.2, 1 - 0.2) = (4.3, 0.8). So approximately (4.3, 0.8). Let me see where that is in relation to the quadrilateral. Point A is (0,0), B is (2,4), C(7,4), D(7,-2). Point E is near (4.3, 0.8), which is inside the quadrilateral. Let me check the line AE. From A(0,0) to E(4.3, 0.8). Let me calculate the angle between AE and AB, and AE and AD.

Vector AE is (4.3, 0.8). Let's compute angle between AE and AB.

Vector AB is (2,4). The angle between AE and AB:

Dot product: 4.3*2 + 0.8*4 = 8.6 + 3.2 = 11.8

|AE| = sqrt(4.3² + 0.8²) ≈ sqrt(18.49 + 0.64) = sqrt(19.13) ≈ 4.373

|AB| = sqrt(20) ≈ 4.472

cos(theta) = 11.8 / (4.373 * 4.472) ≈ 11.8 / 19.56 ≈ 0.603

Therefore theta ≈ arccos(0.603) ≈ 53°.

Similarly, angle between AE and AD: vector AD is (7, -2). Vector AE is (4.3, 0.8). Dot product: 4.3*7 + 0.8*(-2) = 30.1 - 1.6 = 28.5

|AE| ≈ 4.373, |AD| ≈ 7.28

cos(theta) = 28.5 / (4.373 * 7.28) ≈ 28.5 / 31.83 ≈ 0.895

Therefore theta ≈ arccos(0.895) ≈ 26.5°

Therefore, angle between AE and AB is ≈53°, and angle between AE and AD is ≈26.5°, which suggests that AE is the second trisector of angle BAD, since angle BAD is ≈79.5°, divided into three parts of ≈26.5°, so the first trisector is at 26.5° from AB, the second trisector at 26.5° + 26.5° = 53° from AB, which matches the angle between AE and AB of ≈53°. Therefore, AE is the second trisector, dividing angle BAD into 26.5° and 53°, but since the total angle is 79.5°, that would mean that from AB to AE is 53°, and from AE to AD is 26.5°, which is inconsistent with the trisection. Wait, no: if angle BAD is trisected into three equal parts, then from AB to first trisector is x, first to second trisector is x, and second to AD is x. So total 3x. If x ≈26.5°, then the angle from AB to first trisector is 26.5°, then first to second trisector another 26.5°, totaling 53°, then from second trisector to AD is another 26.5°. Therefore, if AE is the second trisector, then angle between AE and AD is 26.5°, which matches our calculation (≈26.5°). And angle between AE and AB is 53°, which is 2x. So AE is the second trisector, dividing angle BAD into 2x and x.

Similarly, AF is another trisector? But in the diagram, there is also AF drawn. Wait, the Asymptote code draws A--E--D and A--F--D. So AF is another line from A to F. Let's analyze AF.

Point F is at (5,3). Let me calculate angles for AF.

As before, angle between AF and AB is ≈32.5°, and angle between AF and AD is ≈47°. Let me see if that corresponds to any trisection.

If angle BAD is trisected, then possible angles from AB are 26.5°, 53°, and 79.5° (which is AD). If angle between AF and AB is ≈32.5°, which is between the first trisector (26.5°) and second trisector (53°), so AF is not a trisector. However, the problem says that angle BAD is trisected, so there must be lines indicating the trisectors. In the diagram, only AE is drawn as a trisector? Or AF?

Wait, the Asymptote code draws A--E--D and A--F--D. So both AE and AF are drawn from A to E and F, which are connected to D. So maybe AE and AF are both trisectors? But there are two trisectors for angle BAD, so AE and AF could be those two trisectors. Let me check.

If AE is the second trisector (angle from AB to AE is 2x ≈53°), then AF must be the first trisector, with angle from AB to AF being x ≈26.5°. Let me check if AF makes angle ≈26.5° with AB.

Earlier, we calculated angle between AF and AB as ≈32.5°, which is not x ≈26.5°. So this is conflicting.

Wait, unless my estimation is off. Let me check again.

If angle between AF and AB is ≈32.5°, and angle between AE and AB is ≈53°, and angle between AE and AD is ≈26.5°, then if AF is between AB and AE, it might be the first trisector. But according to the angle calculations, it's not matching.

Alternatively, maybe there's a miscalculation. Let me recalculate the angle between AF and AB.

Vector AF is (5,3). Vector AB is (2,4). Their dot product is 5*2 + 3*4 = 10 + 12 = 22. |AF| = sqrt(5² + 3²) = sqrt(34) ≈ 5.830. |AB| = sqrt(2² + 4²) = sqrt(20) ≈ 4.472. So cos(theta) = 22 / (5.830 * 4.472) ≈ 22 / 26.07 ≈ 0.844. The arccos of 0.844 is approximately 32.5°. Let me verify with a calculator. If I take arccos(0.844), it's approximately 32.5°, yes. So that seems correct.

Now, if the first trisector makes an angle of x ≈26.5° with AB, and AF is making 32.5°, which is larger, then AF is not a trisector. This suggests that maybe the diagram's labels are different. Perhaps the x's are not the trisection angles but something else? Or maybe my assumption that angle BAD is 3x is incorrect?

Wait, maybe the trisection is not of the entire angle, but of some other angle? No, the problem says angle BAD and angle CDA are trisected.

Alternatively, maybe the diagram is not drawn to scale, and the coordinates are just for drawing purposes, and the actual angles are different. But the Asymptote code is provided, and the labels of x and y, and the angles 110° and 100°, so maybe I need to use those.

Wait, let's try another approach. Let me use the given angles at B and C, and the fact that quadrilateral ABCD has angles at B = 110°, at C = 100°, and we have angles at A and D to find.

From earlier, we have angle A + angle D = 360 - 110 - 100 = 150°. So angle A + angle D = 150°. If angle BAD is 3x and angle CDA is 3y, then 3x + 3y = 150 => x + y = 50°, as before.

Now, if I can relate angle AFD to x and y, I need to find relations in triangle AFD or other triangles.

Let me consider triangle AFD. To find angle AFD, I need to know two other angles. Let's consider angles at A and D.

Angle at A in triangle AFD is angle FAD. Angle at D in triangle AFD is angle FDA.

If I can express these angles in terms of x and y, then I can find angle AFD.

Assuming that AF is a trisector of angle BAD, then angle FAD is either x or 2x. Similarly, if FD is a trisector of angle CDA, then angle FDA is y or 2y.

But how to determine which?

Let me consider the Asymptote code labels. There are three x's and three y's. The x's are at positions that might correspond to the three trisection angles of angle BAD. Similarly for y. If angle BAD is trisected into three angles, each labeled x, then the sum of those three angles is 3x = angle BAD. Similarly for angle CDA.

If there are three x's, perhaps each of the trisection angles is labeled x. For example, between AB and first trisector: x, between first and second trisectors: x, between second and AD: x. Similarly for y.

Now, in the diagram, there are two lines from A: AE and AF. If these are the two trisectors, then one of them is the first trisector and the other is the second trisector.

If I assume that AE is the second trisector (as we saw earlier, it makes angle ≈53° with AB, which is 2x where x ≈26.5°), and AF is the first trisector, making angle x ≈26.5° with AB. But earlier calculation suggests that AF makes angle ≈32.5° with AB, which is not x ≈26.5°. However, maybe there's a calculation error?

Wait, let me recalculate the angle between AF and AB. Maybe I made a mistake in the coordinates of F?

Wait, in the Asymptote code, F is at (5, 3). Let me confirm that. The Asymptote code says:

pair E, F;
E = (4.5-.2,1-.2); F = (5, 3);

So E is at (4.5 - 0.2, 1 - 0.2) = (4.3, 0.8), and F is at (5,3). So F is at (5, 3). Let me check the angle between AF and AB again.

Vector AF is from A(0,0) to F(5,3): (5,3). Vector AB is from A(0,0) to B(2,4): (2,4).

The angle between (5,3) and (2,4). Let me calculate this angle using the tangent formula.

The slope of AF is 3/5 = 0.6, and the slope of AB is 4/2 = 2. The angle between them can be calculated as:

tan(theta) = |(m2 - m1)/(1 + m1*m2)| = |(2 - 0.6)/(1 + (2)(0.6))| = |1.4 / 2.2| ≈ 0.6364

Therefore theta ≈ arctan(0.6364) ≈ 32.5°, which confirms the previous result. So angle between AF and AB is ≈32.5°, which is greater than x ≈26.5°, suggesting AF is not the first trisector.

But if AE is the second trisector, then angle between AE and AB is ≈53°, which is 2x, and angle between AE and AD is x ≈26.5°, as calculated earlier. So if AE is the second trisector, then the first trisector would be somewhere between AB and AE, making x ≈26.5° with AB. Let's call this first trisector line AG, where G is some point. If the first trisector is AG, then angle between AB and AG is x ≈26.5°, and between AG and AE is x ≈26.5°, and between AE and AD is x ≈26.5°.

But in the diagram, there is no line AG, only AE and AF. So perhaps AF is a different line? Maybe it's a trisector of angle CDA?

Wait, the problem says angle BAD and angle CDA are trisected. So both angles are trisected, with two trisectors each. In the diagram, from A, there are two lines: AE and AF. From D, there might be two trisectors, but only one is shown, FD. So FD is a trisector of angle CDA.

If FD is a trisector of angle CDA, then let's calculate the angle between DC and DF, and between DF and DA.

Point F is at (5,3). Let me calculate the angle between DF and DC, and between DF and DA.

Vector DF is from D(7,-2) to F(5,3): (5-7, 3 - (-2)) = (-2, 5).

Vector DC is from D(7,-2) to C(7,4): (0,6).

Vector DA is from D(7,-2) to A(0,0): (-7, 2).

First, let's find the angle between DF and DC.

Vector DF = (-2,5), vector DC = (0,6).

Dot product: (-2)(0) + (5)(6) = 30.

|DF| = sqrt(4 + 25) = sqrt(29) ≈ 5.385

|DC| = 6

cos(theta) = 30 / (5.385 * 6) ≈ 30 / 32.31 ≈ 0.9287

So theta ≈ arccos(0.9287) ≈ 21.8°

Similarly, angle between DF and DA: vector DF = (-2,5), vector DA = (-7,2).

Dot product: (-2)(-7) + (5)(2) = 14 + 10 = 24.

|DF| ≈ 5.385, |DA| ≈ 7.28

cos(theta) = 24 / (5.385 * 7.28) ≈ 24 / 39.1 ≈ 0.614

Therefore theta ≈ arccos(0.614) ≈ 52°.

So angle between DF and DC is ≈21.8°, and between DF and DA is ≈52°. The total angle CDA is ≈74°, so 21.8 + 52 ≈73.8°, which matches. Therefore, if FD is a trisector of angle CDA, then angle between DC and DF is y, and between DF and DA is 2y. So y ≈21.8°, and 2y ≈43.6°, but we calculated 52°, which is inconsistent. Alternatively, if FD is dividing angle CDA into y and 2y, then 3y = 74°, so y ≈24.7°, but our calculation gave angle between DC and DF as ≈21.8°, and between DF and DA as ≈52°, which is not a multiple of y. This inconsistency suggests that FD is not a trisector, or my calculations are wrong.

Wait, but if angle CDA is trisected, then it should be divided into three equal parts. If angle CDA is 74°, then each trisection angle is ≈24.7°. Let me check if FD divides it into 24.7° and 49.4°, but our calculation showed 21.8° and 52°, which doesn't align. So something is wrong here.

Alternatively, perhaps the labels in the Asymptote code are not corresponding to the trisection angles but to other parts. Given the confusion, perhaps I should use the earlier relation x + y = 50° and assume that in triangle AFD, angle AFD = 180 - 2x - 2y = 80°, regardless of the exact positions, since x + y = 50°.

But why would angle AFD be 180 - 2x - 2y? If in triangle AFD, angle at A is 2x and angle at D is 2y, then yes. But how do I know those are the correct angles?

Alternatively, let me think about the problem differently. Since angle BAD is trisected, let me call the trisectors as follows: from A, draw two lines, dividing angle BAD into three equal parts. Let's call the first trisector as line AM, making angle x with AB, and second trisector as line AN, making 2x with AB. Similarly, for angle CDA, trisectors: from D, draw two lines, dividing angle CDA into three equal parts. Let's call the first trisector as line DP, making angle y with DC, and second trisector as line DQ, making 2y with DC.

Now, if in the diagram, the lines drawn are AM and AN from A, and DP and DQ from D. But in the Asymptote code, only lines from A to E and A to F, and from F to D. So perhaps F is the intersection of two trisectors? Or maybe F is on a trisector of angle CDA?

If F is on the trisector of angle CDA, then DF would be that trisector, and AF would be some line. But without more information, it's hard to say.

Alternatively, using the given labels in the diagram, with x and y, and given that x + y = 50°, and looking for angle AFD, which is likely 80° as previously calculated, and since this is a common angle in such trisection problems, I might conclude that the answer is 80°.

However, to confirm, let me think about the diagram's angles. Let me assume that angle AFD is 80°, and see if it makes sense with the other angles.

If x + y = 50°, and angle AFD = 180 - 2(x + y) = 80°, this is a solid relation, assuming that angle FAD = 2x and angle FDA = 2y.

Alternatively, if angle FAD = x and angle FDA = y, then angle AFD = 180 - x - y = 130°, but this would require that AF is a trisector of angle BAD and FD is a trisector of angle CDA, but based on the earlier calculations, it seems that AF is not a trisector. However, if this were the case, then angle AFD = 130°, but I need to check if this is possible.

Wait, let me think about the given labels. The Asymptote code labels three x's and three y's. If angle BAD is trisected into three x's, then each trisection angle is x, so x = angle BAD / 3. Similarly for y. If the three x's are the trisection angles, then in triangle AFD, if angle FAD is x and angle FDA is y, then angle AFD = 180 - x - y. But since x + y = 50°, then angle AFD = 130°. But why would angle FAD be x and angle FDA be y?

If AF is the first trisector of angle BAD, then angle FAD = 2x (since from AF to AD is x, and from AB to AF is x, total angle BAD = 3x). Wait, if AF is the first trisector, then angle between AB and AF is x, and between AF and AD is 2x. So angle FAD is 2x. Similarly, if FD is the first trisector of angle CDA, then angle between DC and FD is y, and between FD and DA is 2y. So angle FDA is 2y. Hence, angle AFD = 180 - 2x - 2y = 80°, as before.

But if AF is the second trisector, then angle FAD = x, and if FD is the second trisector, angle FDA = y, then angle AFD = 180 - x - y = 130°. But without knowing which trisectors AF and FD are, it's hard to say.

However, given that there are three x's and three y's in the diagram, it's more likely that each trisection angle is labeled, and therefore, angle FAD and angle FDA are multiples of x and y. Since x + y = 50°, and the only way to get a unique answer is if angle AFD is 80°, which is 180 - 2(x + y), I think this is the intended solution.

Therefore, I will go with angle AFD = 80°, so the degree measure is 80 degrees.

**Final Answer**
\boxed{80}

To determine the measure of angle $ \angle AFD $ in quadrilateral $ ABCD $ with trisected angles $ \angle BAD $ and $ \angle CDA $, we proceed step-by-step using geometric reasoning and algebraic relations.

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### **1. Understanding the Geometry**

We are given that:

- Quadrilateral $ ABCD $ includes angles $ \angle BAD $ and $ \angle CDA $, which are each **trisected**.
- Points $ E $ and $ F $ lie on the trisectors from $ A $ and $ D $, respectively.
- Diagram indicates that lines $ AE $ and $ AF $ originate from vertex $ A $, while $ DF $ extends from $ D $.
- Labeling in the diagram suggests that the trisectors split $ \angle BAD $ into three equal parts, each of measure $ x $, and $ \angle CDA $ into three equal parts, each of measure $ y $.

From this, we can assign:

- $ \angle BAD = 3x $
- $ \angle CDA = 3y $
- Angles at $ B $ and $ C $ are given as $ 110^\circ $ and $ 100^\circ $, respectively.

---

### **2. Using the Angle Sum of a Quadrilateral**

The sum of the interior angles in any quadrilateral is:

$$
\text{Sum of angles} = 360^\circ
$$

Thus:

$$
\angle BAD + \angle CDA + \angle ABC + \angle BCD = 360^\circ
$$

$$
3x + 3y + 110^\circ + 100^\circ = 360^\circ
\Rightarrow 3x + 3y = 150^\circ
\Rightarrow x + y = 50^\circ
$$

This is our key relationship:
$$
x + y = 50^\circ
$$

---

### **3. Analyzing Triangle $ AFD $**

We are to find $ \angle AFD $, the angle at vertex $ F $ in triangle $ AFD $. Let’s express this angle using known quantities.

In triangle $ AFD $, the three angles are:

- $ \angle FAD $: the angle at vertex $ A $, formed by the trisection of $ \angle BAD $. Depending on which trisector corresponds to ray $ AF $, $ \angle FAD = 2x $ (if $ AF $ is the second trisector from $ A $).
- $ \angle FDA $: the angle at vertex $ D $, formed by the trisection of $ \angle CDA $. Similarly, if $ FD $ is the second trisector from $ D $, $ \angle FDA = 2y $.

Thus, the angle at $ F $ is:

$$
\angle AFD = 180^\circ - \angle FAD - \angle FDA = 180^\circ - 2x - 2y
$$

Now, substituting $ x + y = 50^\circ $:

$$
\angle AFD = 180^\circ - 2(x + y) = 180^\circ - 2(50^\circ) = 180^\circ - 100^\circ = \boxed{80^\circ}
$$

---

### **4. Final Result**

Thus, the degree measure of angle $ \angle AFD $ is:

$$
\boxed{80}
$$

[Assistant]
Explanation:
Okay, so I need to find the measure of angle AFD in quadrilateral ABCD. The problem says that angles BAD and CDA are trisected, and there are some labels with x and y, as well as given angles of 110° and 100°. Let me try to visualize the diagram based on the Asymptote code provided.

First, let me sketch quadrilateral ABCD mentally. Points A, B, C, D are given with coordinates: A is at (0,0), B at (2,4), C at (7,4), and D at (7,-2). So, connecting A to B to C to D and back to A. The Asymptote code draws A to B to C to D to A. Then there are points E and F, with E near (4.5,1) and F at (5,3). The lines A-E-D and A-F-D are drawn, so E and F are points on the trisectors of angles BAD and CDA, respectively.

The labels mention angles labeled with x and y, and specific angles of 110° and 100°. The problem statement says angle BAD and angle CDA are trisected. Let me note that. So angle BAD is trisected into three equal parts, and angle CDA is also trisected into three equal parts. The Asymptote code labels some parts with x and y. Let me try to figure out how they are placed.

Looking at the Asymptote code:

Label "x" at (1, 1.5) and (2, 1), (2, -1). The label "y" at (5.5, 0.5), (6.5, 0), (5.5, -1.5). Then labels "110°" near point B and "100°" near point C. Wait, the Asymptote code says:

label("$110^{\circ}$",(2.5,3.5)); label("$100^{\circ}$",(6.5-.2,3.5));

So, those labels seem to be near the angles at B and C. Wait, the coordinates (2.5,3.5) is near point B which is at (2,4), so maybe angle at B is 110°, and angle at C is 100°? Wait, but angle at C is between points B, C, D. Wait, but the label for 100° is at (6.5 - 0.2, 3.5) which is (6.3, 3.5). Since point C is at (7,4), so that's near point C. So maybe angle at B is 110°, angle at C is 100°? But the problem didn't specify those angles, but they are given in the Asymptote code. Wait, but maybe the original problem has those angles, but the Asymptote code is just an illustration? Wait, the problem says "angle BAD and angle CDA are trisected as shown." So perhaps the diagram is drawn with specific angles, and the labels x, y, and 110°, 100° are part of the diagram. Maybe angle BAD is trisected into three angles, each labeled x, and angle CDA is trisected into three angles labeled y? Wait, looking at the Asymptote code:

label("$x$", (1, 1.5), S); label("$x$", (2, 1), S+W); label("$x$", (2, -1), N+N+N+W);

So there are three x's. Let me see their coordinates. The first x is at (1, 1.5), which is near the middle of AB? Because A is at (0,0), B is at (2,4). So (1,1.5) is somewhere between A and B? Wait, but angle BAD is at point A between BA and DA. Wait, angle BAD is the angle at A between AB and AD. Similarly, angle CDA is the angle at D between DC and DA.

So, if angles BAD and CDA are trisected, then angle BAD is divided into three equal angles, and angle CDA is divided into three equal angles. The Asymptote code draws lines from A to E and from A to F. Wait, but E is connected to D, so maybe E is on the trisector of angle BAD? Similarly, F is connected to D, so F is on the trisector of angle CDA? Or maybe E is on the trisectors of angle BAD and F is on trisectors of angle CDA?

Wait, the Asymptote code says:

draw(A--E--D); draw(A--F--D);

So E is connected to A and D, so E is a point inside the quadrilateral connected to A and D. Similarly, F is connected to A and D. Wait, but A-F-D would be a line from A to F to D? But then if F is on the trisector of angle CDA, maybe?

Wait, maybe E is one of the trisectors of angle BAD and F is one of the trisectors of angle CDA? Let me think.

First, let me figure out the angles at A and D. The problem mentions trisection, so maybe angle BAD is trisected into three angles, each of measure x, and angle CDA is trisected into three angles, each of measure y. Then the labels x and y might be those trisected angles? But the Asymptote code labels three x's and three y's. Let me try to see.

But maybe looking at coordinates can help. Let me compute coordinates of points to figure out angles.

Coordinates:

A = (0,0)

B = (2,4)

C = (7,4)

D = (7,-2)

So AB is from (0,0) to (2,4). AD is from (0,0) to (7,-2). So angle BAD is the angle between AB and AD.

Similarly, angle CDA is the angle at D between DC and DA. DC is from D to C, which is from (7,-2) to (7,4), so that's straight up along x=7. DA is from D to A, which is from (7,-2) to (0,0), so that's a line going to the left and up.

So angle CDA is between DC (upwards vertical) and DA (line from D to A). Let me try to compute angles BAD and CDA first.

But maybe it's better to find angle BAD first. Let me calculate vectors.

For angle BAD at point A between AB and AD.

Vector AB is (2,4) - (0,0) = (2,4)

Vector AD is (7,-2) - (0,0) = (7,-2)

So angle between AB and AD can be calculated using the dot product.

Dot product of AB and AD is 2*7 + 4*(-2) = 14 - 8 = 6

The magnitude of AB is sqrt(2^2 + 4^2) = sqrt(4 + 16) = sqrt(20) = 2*sqrt(5)

The magnitude of AD is sqrt(7^2 + (-2)^2) = sqrt(49 + 4) = sqrt(53)

So cos(theta) = 6 / (2*sqrt(5)*sqrt(53)) = 6 / (2*sqrt(265)) = 3 / sqrt(265)

But maybe that's not necessary. Wait, maybe we can find angle BAD?

Alternatively, perhaps using coordinates to find the slopes and compute the angle.

Slope of AB: from A(0,0) to B(2,4): slope is (4-0)/(2-0) = 2

Slope of AD: from A(0,0) to D(7,-2): slope is (-2 - 0)/(7 - 0) = -2/7

So angle between two lines with slopes m1 and m2 is given by:

tan(theta) = |(m2 - m1)/(1 + m1*m2)|

So tan(theta) = |(-2/7 - 2)/(1 + (2)*(-2/7))| = |(-2/7 - 14/7)/(1 - 4/7)| = |(-16/7)/(3/7)| = | -16/3 | = 16/3

So theta is arctangent of 16/3. Let me compute that. Well, arctan(16/3) is some angle. Let me note that 16/3 ≈ 5.333, so arctan(5.333) is roughly 79 degrees? Wait, tan(80 degrees) is approx 5.671, so arctan(16/3) ≈ 79 degrees? Wait, tan(79°) ≈ 5.145, so 16/3 is about 5.333, so angle is a bit more. Maybe around 79.5 degrees? Let me check:

tan(79°) ≈ 5.145

tan(79.5°): let's compute:

Using calculator, tan(79.5 degrees) ≈ tan(79 + 0.5) ≈ ?

But maybe I can calculate it. Alternatively, maybe I can find exact value? Not sure. Maybe it's not necessary right now.

Similarly, angle CDA. Let me compute angle at D between DC and DA.

Vector DC is from D to C: (7,4) - (7,-2) = (0,6). So vector DC is straight up along positive y-axis.

Vector DA is from D to A: (0,0) - (7,-2) = (-7, 2). So vector DA is (-7, 2). So angle between DC (0,6) and DA (-7, 2).

So, angle between vector DC (0,6) and DA (-7,2). Let me calculate that angle.

First, vector DC is (0,6), direction is straight up. Vector DA is (-7,2). The angle between DC and DA can be calculated using dot product.

Dot product: (0)(-7) + (6)(2) = 12

Magnitude of DC is 6.

Magnitude of DA is sqrt( (-7)^2 + 2^2 ) = sqrt(49 +4) = sqrt(53)

So cos(theta) = 12 / (6 * sqrt(53)) ) = 2 / sqrt(53)

Therefore theta = arccos(2 / sqrt(53)) ≈ arccos(0.277). Let me compute that. sqrt(53) ≈ 7.28, so 2 / 7.28 ≈ 0.275. arccos(0.275) ≈ 73.7 degrees. Hmm, so angle CDA is approximately 73.7 degrees. But maybe exact value? Let me think.

Wait, but maybe the problem does not require us to compute these angles directly. Instead, given that angles BAD and CDA are trisected, and there are some labels of x and y, as well as given angles of 110° and 100°, which are probably angles at B and C? The Asymptote code labels "110°" at (2.5,3.5) and "100°" at (6.5 - 0.2, 3.5). Given that point B is at (2,4), so (2.5,3.5) is near B, so that's likely angle at B. Similarly, point C is at (7,4), so label "100°" is near C, so angle at C is 100°, and angle at B is 110°? Wait, but in quadrilaterals, the sum of angles is 360°, so if we can find other angles, we can check.

But let me check the given data. So quadrilateral ABCD. Angles at B and C are given as 110° and 100°, and angles at A and D are being trisected. If I can find angles at A and D, then maybe I can find the trisection angles x and y, then find angle AFD.

Alternatively, since angle BAD is trisected into three x's and angle CDA is trisected into three y's. So angle BAD = 3x, angle CDA = 3y. Then, in quadrilateral ABCD, the sum of internal angles is 360°, so angle A + angle B + angle C + angle D = 360°. If we can write that equation with angles A = 3x, angle D = 3y, angle B = 110°, angle C = 100°, then 3x + 110 + 100 + 3y = 360. Therefore, 3x + 3y = 360 - 210 = 150. So x + y = 50°. So, x + y = 50°. That's one equation.

But we need more equations to find x and y. However, the Asymptote code labels x's and y's. Let me see. The diagram labels three x's and three y's. Let me think about how the trisection occurs. So angle BAD is trisected into three angles, each x. So if angle BAD is at vertex A, between AB and AD, then the trisectors would divide that angle into three equal parts. Similarly, angle CDA is at vertex D, between DC and DA, trisected into three angles each y. So if I can find relations based on the diagram, maybe involving triangle AFD or other triangles.

Alternatively, perhaps using the fact that lines AE and AF are trisectors. Wait, in the Asymptote code, they draw A--E--D and A--F--D. So E is connected to A and D, and F is connected to A and D. So AE and AF are trisectors of angle BAD and angle CDA?

Wait, angle BAD is at A, so if AE is one of the trisectors, then angle between AB and AE is x, angle between AE and AF is x, and angle between AF and AD is x. Similarly, for angle CDA at D, trisected into three angles y, so the trisectors would be from D, but the diagram shows lines from A to F to D. Wait, unless F is a point on the trisector of angle CDA, meaning that DF is a trisector?

But the Asymptote code draws A--F--D. So point F is connected to A and D. So maybe F is on the trisector of angle CDA? Wait, angle CDA is at D, so the trisectors would be from D. However, the diagram shows a line from A to F to D. Therefore, maybe AF is a trisector of angle BAD, and FD is a trisector of angle CDA? But that would mean FD is a trisector of angle CDA. But angle CDA is at D, so trisectors would be lines from D. Therefore, maybe FD is a trisector of angle CDA, meaning that DF divides angle CDA into y and 2y? Or something like that?

Wait, this is getting confusing. Let me try again. Let me think about trisectors.

Angle BAD is at A, between BA and DA. If it is trisected, then two trisectors are drawn, dividing angle BAD into three equal angles. Similarly, angle CDA is at D, between CD and DA, trisected into three equal angles.

In the diagram, they have lines from A to E and A to F. So maybe E is one trisector of angle BAD, and F is one trisector of angle CDA? Wait, but if F is a trisector of angle CDA, which is at D, then the line DF would be the trisector. But the diagram shows A-F-D. Therefore, AF is a line from A to F, which is connected to D. Therefore, AF is not a trisector of angle CDA, but perhaps of angle BAD?

Wait, angle BAD is at A, so trisectors would be lines from A. Therefore, AE and AF are trisectors of angle BAD? Then there would be two trisectors, dividing BAD into three parts. Similarly, angle CDA is at D, so trisectors would be lines from D. But in the diagram, the lines from D would be... the diagram shows A-F-D, so FD is connected to D, which could be a trisector. So FD is a trisector of angle CDA. So F is a point such that DF trisects angle CDA. So then AF connects to F, which is somewhere inside the quadrilateral.

Alternatively, maybe E is a trisector of angle BAD and F is a trisector of angle CDA. So AE is a trisector of angle BAD, and DF is a trisector of angle CDA. But the diagram has A-E-D and A-F-D. So E is between A and D, connected to A and D? Wait, if E is on the trisector of angle BAD, then it's a line from A going into the quadrilateral, but E is connected to D? So E is connected to A and D, so it's a line from E to D. Therefore, E lies somewhere such that AE is a trisector of angle BAD, and ED is a line from E to D. Similarly for F.

This is getting too vague. Maybe I need to use the given angles of 110° and 100°, and the x and y labels. Let me think again about the earlier equation: x + y = 50°, from angle sum of quadrilateral. If I can find another equation relating x and y, I can solve for x and y, and then use that to find angle AFD.

Alternatively, maybe triangle AFD is involved. Let me think about triangle AFD. To find angle AFD, we need to know something about triangle AFD. If I can find the measures of other angles in triangle AFD or some relations, maybe using the trisection angles.

But how?

Alternatively, using the given labels of x and y. The Asymptote code labels three x's and three y's. Let me try to figure out where they are placed. The first label for x is at (1,1.5), which is near AB. Second x at (2,1) S+W, which is near the middle of AB and AD? Third x at (2,-1), N+N+N+W, which is near AD? Similarly, y's are at (5.5+.3, .5-.3), which is near DC and DA. Maybe these labels correspond to the trisection angles?

Let me think. For angle BAD, which is at A, if it's trisected into three angles of x each, then maybe the labels x are on those trisected angles. Similarly for angle CDA trisected into three angles of y each.

But in the diagram, the Asymptote code draws A to E to D and A to F to D. So E is connected to D, which is part of angle CDA? Or is E connected to D as part of angle BAD?

Wait, if angle BAD is trisected, then the trisectors would be two lines from A dividing angle BAD into three equal parts. Similarly, angle CDA is trisected, so two lines from D dividing angle CDA into three equal parts. However, in the diagram, there are two lines from A: AE and AF, and since E and F are connected to D, that suggests that E and F are points on those trisectors. Wait, but if angle BAD is trisected, then the trisectors would be two lines from A, but in the diagram, they have two lines from A: AE and AF. If AE and AF are the trisectors of angle BAD, then they divide angle BAD into three parts. Similarly, if there is another line from D for angle CDA.

But the diagram shows A-F-D. If F is connected to D, then maybe DF is one of the trisectors of angle CDA. Therefore, DF divides angle CDA into two parts, one being y and the other being 2y or something? Wait, but if angle CDA is trisected, then there should be two trisectors dividing it into three equal angles. So if DF is one of the trisectors, then the angle between DC and DF is y, and between DF and DA is 2y? Or perhaps between DC and DF is y, DF to DA is y, and then another trisector? Wait, no, if it's trisected, there are two trisectors, dividing the angle into three equal parts. So angle between DC and first trisector is y, between the two trisectors is y, and between the second trisector and DA is y. Therefore, the total angle CDA is 3y.

But in the diagram, the line from D is DF. If DF is one of the trisectors, then maybe FD is the first trisector, dividing angle CDA into y and 2y, but without more information, it's hard to tell.

Alternatively, since the diagram labels x and y in certain locations, perhaps those x's are the trisected angles at A, and y's at D. Let me think. The three x's are at (1,1.5), (2,1), and (2,-1). The (1,1.5) is near AB, so maybe that is one of the trisection angles between AB and AE? Then the next x at (2,1) is near AB and AD? Wait, this is too vague. Maybe I need to consider that the angles labeled x are adjacent to AB and AD? Let me try to imagine the diagram again.

Since I can't see the image, but based on Asymptote code, maybe the Asymptote code draws A to E to D, so E is inside the quadrilateral, connected to A and D. Similarly, F is connected to A and D. The labels x are on three different positions. Let me try to think again about the angles.

Given that angle BAD is trisected into three x's, so if angle BAD is 3x, and angle CDA is 3y, and from earlier, x + y = 50°. So if I can find another relation between x and y, I can solve for them. How?

Perhaps using triangle AFD. If I can find angles at A and D related to triangle AFD, then I can find angle at F.

But angle at A for triangle AFD is angle FAD, which is part of angle BAD. Similarly, angle at D for triangle AFD is angle FDA, which is part of angle CDA. Depending on how AE and AF are placed, those angles could be x or multiples thereof.

Wait, assuming that AF is a trisector of angle BAD, then angle FAD would be x or 2x? Similarly, if FD is a trisector of angle CDA, then angle FDA would be y or 2y?

But without knowing exactly which trisectors AE and AF are, it's hard. However, the Asymptote code labels three x's and three y's. If there are three x's, that suggests that angle BAD is divided into three equal parts, each x. Similarly for angle CDA with three y's. So if angle BAD = 3x, angle CDA = 3y. Then, from quadrilateral angle sum, 3x + 3y + 110 + 100 = 360 => 3x + 3y = 150 => x + y = 50° as before.

But how can I find another equation?

Maybe looking at triangle AFD. In triangle AFD, the sum of angles is 180°, so angle FAD + angle FDA + angle AFD = 180°. If I can find angle FAD and angle FDA in terms of x and y, then I can find angle AFD.

If AF is a trisector of angle BAD, then angle FAD could be 2x or x, depending on which trisector it is. Similarly, if FD is a trisector of angle CDA, then angle FDA could be 2y or y.

But the diagram has two lines from A: AE and AF. If AE and AF are the two trisectors of angle BAD, then they divide angle BAD into three x's. So if AF is the first trisector, then angle between AB and AF is x, and between AF and AE is x, and between AE and AD is x. But then the Asymptote code draws A--E--D and A--F--D. So if AF is a trisector, then maybe angle between AF and AD is x? Wait, if the trisection is from AB to AD, then the first trisector would divide angle BAD into x and 2x. The second trisector divides into 2x and x. So depending on which trisector AF is, angle FAD could be 2x or x. Similarly for FD.

But how can I determine?

Alternatively, looking at the labels in the diagram. The Asymptote code labels three x's. If angle BAD is trisected into three angles, each adjacent to AB, then perhaps the angles labeled x are between AB and AE, AE and AF, and AF and AD? But the diagram has three x's. So if AE and AF are the two trisectors, then there are three angles: AB to AE (x), AE to AF (x), AF to AD (x). Similarly for angle CDA.

But in the Asymptote code, the labels for x are at (1,1.5), (2,1), (2,-1). Let me try to think where those could be. The point (1,1.5) is between A(0,0) and B(2,4), and closer to A. The point (2,1) is between A and B? Or is it between AB and AD? Wait, maybe (2,1) is near the middle of AB and AD? Wait, AB goes from (0,0) to (2,4). AD goes from (0,0) to (7,-2). The point (2,1) is somewhere in the middle of the quadrilateral. Maybe it's the intersection of some lines?

Alternatively, maybe the three x's are the three angles formed by the trisection of angle BAD. So, if angle BAD is trisected, there are two lines from A dividing it into three equal parts. Then, each of those angles is labeled x, and those labels are at positions corresponding to those angles.

Similarly, three y's for angle CDA.

But without seeing the diagram, it's challenging, but maybe I can think of the trisection angles.

Alternatively, maybe using the given angles at B and C. The problem mentions "as shown," and in the Asymptote code, the angles at B is 110°, angle at C is 100°, so maybe we can compute angles at A and D?

Earlier, I calculated angle BAD ≈ arctan(16/3) ≈ 79 degrees? Wait, but if angle BAD is trisected, and angle CDA is trisected, but we have from quadrilateral sum that 3x + 3y = 150, so x + y = 50. If I could compute angles at A and D from coordinates, then I can find x and y.

Let me try that. Earlier, I tried computing angle BAD:

Vectors AB = (2,4), AD = (7,-2). The angle between them was calculated using tan(theta) = 16/3 ≈ 5.333, which gave theta ≈ arctan(5.333). Let me compute this angle more accurately.

Let me recall that tan(theta) = 16/3. So theta is arctan(16/3). Let me convert this to degrees.

Using calculator input:

arctan(16/3) ≈ arctan(5.333) ≈ 79.3 degrees.

Similarly, angle CDA was calculated as arccos(2 / sqrt(53)) ≈ arccos(0.277) ≈ 73.7 degrees.

So adding up, angle A ≈ 79.3°, angle B = 110°, angle C = 100°, angle D ≈ 73.7°. Let me check if they add up to 360:

79.3 + 110 + 100 + 73.7 ≈ 79.3 + 110 = 189.3; 189.3 + 100 = 289.3; 289.3 + 73.7 = 363°. Hmm, off by 3°, probably due to approximation errors in calculations.

But let me try to compute angle BAD and CDA more accurately.

First, angle BAD:

Vectors AB = (2,4) and AD = (7,-2).

The formula for the angle between two vectors is:

cos(theta) = (AB · AD) / (|AB| |AD|)

AB · AD = 2*7 + 4*(-2) = 14 -8 = 6

|AB| = sqrt(2^2 + 4^2) = sqrt(20) = 2*sqrt(5)

|AD| = sqrt(7^2 + (-2)^2) = sqrt(49 + 4) = sqrt(53)

Thus,

cos(theta) = 6 / (2√5 * √53) = 6 / (2√(265)) = 3 / √265 ≈ 3 / 16.2788 ≈ 0.1844

So theta ≈ arccos(0.1844) ≈ 80 degrees? Let me check:

cos(80°) ≈ 0.1736

cos(79°) ≈ 0.1908

So 0.1844 is between 79° and 80°, closer to 79.5°. Let me compute:

Let me compute cos(79.5°):

Using calculator, cos(79.5°) ≈ cos(79°30') ≈ 0.1822

Hmm, so 0.1844 is slightly higher. Let me use linear approximation.

Between 79° (cos ≈ 0.1908) and 80° (≈ 0.1736). Wait, the value 0.1844 is between 0.1736 and 0.1908? Wait no, 0.1844 is between 0.1736 and 0.1908? Wait, 0.1736 is 80°, 0.1908 is 79°. So cos(theta) = 0.1844 is between 79° and 80°, closer to 80°? Wait, no:

Wait, as theta increases from 0° to 90°, cos(theta) decreases from 1 to 0. So if cos(theta) = 0.1844, which is less than 0.1908 (cos79°), then theta is greater than 79°, since cos(theta) decreases. Therefore, theta is between 79° and 80°, with cos(theta) = 0.1844. Let me find it more precisely.

Let me denote theta = 79° + d, where d is in degrees between 0 and 1. Then cos(theta) = cos(79° + d) = cos(79°)cos(d) - sin(79°)sin(d). But maybe using linear approximation is better.

Let me approximate. Let me take theta ≈ 79.5°, as before, and compute cos(79.5°):

Actually, let me use calculator-like computation:

We can use the formula:

cos(a + b) ≈ cos(a) - b*sin(a) if b is small in radians.

Let me convert 79° to radians: 79° * π/180 ≈ 1.3788 radians.

Let me take a = 79°, and let me compute cos(a) ≈ 0.1908.

Let me let b be the difference in radians so that cos(a + b) = 0.1844.

So:

cos(a + b) ≈ cos(a) - b*sin(a) = 0.1844.

We have:

0.1908 - b*sin(79°) = 0.1844.

sin(79°) ≈ 0.9816.

Therefore:

0.1908 - 0.9816*b = 0.1844

=> 0.1908 - 0.1844 = 0.9816*b

=> 0.0064 = 0.9816*b

=> b ≈ 0.0064 / 0.9816 ≈ 0.0065 radians.

Convert 0.0065 radians to degrees: 0.0065 * (180/π) ≈ 0.0065 * 57.3 ≈ 0.372 degrees.

Therefore theta ≈ 79° + 0.372° ≈ 79.37°, so approximately 79.4°.

Similarly, angle at D: angle CDA. Earlier, we found cos(theta) = 2 / sqrt(53) ≈ 2 / 7.28 ≈ 0.2747. Therefore theta ≈ arccos(0.2747). Let me compute this.

cos(74°) ≈ 0.2756, which is very close to 0.2747. Therefore, angle CDA ≈ 74°, since cos(74°) ≈ 0.2756, and our value is slightly less, so angle is slightly more than 74°, say 74.1° or so.

Therefore, angles at A ≈ 79.4°, angle at D ≈ 74°, angle at B is 110°, angle at C is 100°, sum ≈ 79.4 + 110 + 100 + 74 ≈ 363.4°, which is off by about 3.4°, likely due to approximation errors in angle CDA. But if we take angle BAD ≈ 79.4°, angle CDA ≈ 74°, then 3x ≈ 79.4° => x ≈ 26.5°, 3y ≈ 74° => y ≈ 24.7°, so x + y ≈ 51.2°, which is close to our earlier result of x + y = 50°, considering rounding errors. So approximately x + y = 50°, which seems consistent.

But how does this help me? If I need to find angle AFD. If I can find angles FAD and FDA in triangle AFD, then angle AFD = 180° - angle FAD - angle FDA.

Assuming that AF is a trisector of angle BAD, then angle FAD could be x or 2x depending on which trisector it is. Similarly, if FD is a trisector of angle CDA, then angle FDA could be y or 2y.

But how do I know which trisectors AF and FD are?

Looking back at the Asymptote code: there is a label for x at (1,1.5) which is near AB, and another x at (2,1) which is closer to AD? Maybe those are the trisection angles. Similarly for y. Since there are three x's, it's likely that angle BAD is divided into three equal parts, each labeled x, and similarly for angle CDA.

Assuming that AF is the first trisector from A, so angle between AB and AF is x, then angle FAD would be 2x, but depending on how many trisection lines there are. Wait, angle BAD is trisected into three equal parts, so there are two trisectors. If AF is one trisector, then AF divides angle BAD into x and 2x. Similarly, the other trisector divides it into 2x and x. So if AF is the first trisector, then angle between AB and AF is x, and between AF and AD is 2x. Similarly, if FD is a trisector of angle CDA, then angle between DC and DF is y, and between DF and DA is 2y. Therefore, angle FDA would be 2y.

Therefore, in triangle AFD, angle FAD = 2x, angle FDA = 2y, and angle AFD = 180 - 2x - 2y = 180 - 2(x + y). Since we know x + y = 50°, then angle AFD = 180 - 2*50 = 180 - 100 = 80°. Wait, so angle AFD is 80°? But let me verify if that makes sense.

Alternatively, if angle FAD is x and angle FDA is y, then angle AFD would be 180 - x - y = 180 - 50 = 130°, which doesn't seem right. But why?

But this depends on which trisectors AF and FD are. If AF is dividing angle BAD into x and 2x, and FD divides angle CDA into y and 2y, then angle FAD is 2x and angle FDA is 2y. If that's the case, then angle AFD is 180 - 2x - 2y = 180 - 2*50 = 80°, which would be the answer. But is that correct?

Alternatively, if AF is the second trisector, then angle FAD = x and angle FDA = y, then angle AFD would be 180 - x - y = 130°, which is different. So how do I know which one it is?

Looking back at the Asymptote code, the labels for x's are at positions that may indicate where the trisectors are. For example, label x at (1,1.5): this is near the area between AB and AD? Wait, (1,1.5) is on the line from A(0,0) to B(2,4), closer to A? Wait, the coordinates (1,1.5) is halfway between A(0,0) and (2, 3), which is not exactly on AB. Wait, AB is from (0,0) to (2,4), so parametric equations. Let me check if (1,1.5) is on AB. If we go from A(0,0) to B(2,4), then at t=0.5, the point would be (1, 2). But (1, 1.5) is lower than that. Therefore, (1,1.5) is not on AB. Similarly, (2,1): is that on AD? AD is from (0,0) to (7,-2). Let me check if (2,1) is on AD. The parametric equation for AD is x = 7t, y = -2t for t from 0 to 1. So at t=2/7, x=2, y=-4/7 ≈ -0.57. But (2,1) is not on AD. Similarly, (2,-1): let's check if that's on AD. For y=-1, solve -2t = -1 => t = 0.5, then x=7*0.5=3.5. So (3.5, -1) is on AD. But (2,-1) is not. So these x's are not on the sides, but rather inside the quadrilateral. So perhaps they are labels for angles?

If angle BAD is trisected into three angles, maybe the trisectors are inside the quadrilateral, and the labels x are indicating those angles. So if I imagine angle BAD being trisected, then there are two lines from A dividing it into three equal parts. The diagram shows lines AE and AF. If those are the trisectors, then each trisection angle is x. Suppose that angle between AB and AE is x, angle between AE and AF is x, angle between AF and AD is x. Then, angle FAD is 2x. Similarly, if angle CDA is trisected by FD into three angles y, then the angle between DC and DF is y, between DF and FD... Wait, DF is a trisector. If FD is the first trisector, then angle between DC and FD is y, and between FD and DA is 2y. Therefore, angle FDA is 2y. Thus, angle AFD = 180 - 2x - 2y = 180 - 2(x + y) = 180 - 2*50 = 80°. Therefore, angle AFD is 80 degrees.

But how do I confirm this? Because if the labels x are the trisection angles, then each trisection angle is x, so angle between AB and AE is x, angle between AE and AF is x, angle between AF and AD is x. Similarly, for angle CDA, the trisection angles are y. Therefore, the diagram would have three angles labeled x for angle BAD and three labeled y for angle CDA.

Therefore, assuming that, then in triangle AFD, angle at A is angle FAD = 2x (since from AB to AF is x, then AF to AD is another x, total 2x), and angle at D is angle FDA = 2y (from DC to DF is y, then DF to DA is another y, total 2y). Therefore, angle AFD = 180 - 2x - 2y = 80°, as before.

Alternatively, if angle FAD was x and angle FDA was y, then angle AFD = 180 - x - y = 130°, but that would require that AF is one trisection angle and FD is one trisection angle, which might not be the case.

But given that there are three x's, it seems more likely that angle BAD is divided into three equal parts, each labeled x, and similarly for angle CDA with three y's. Therefore, the angle FAD would be 2x and angle FDA would be 2y, leading to angle AFD = 80°. Let me check if this answer makes sense with the diagram.

Alternatively, maybe there's another approach using the given angles at B and C. Since we know angles at B and C, maybe we can find the other angles.

Wait, earlier approximate calculations suggested angle BAD ≈ 79.4°, angle CDA ≈ 74°, so if angle BAD is 3x ≈ 79.4°, then x ≈ 26.5°, and angle CDA is 3y ≈ 74°, y ≈ 24.7°, so angle AFD = 180 - 2x - 2y ≈ 180 - 2*(26.5 + 24.7) = 180 - 2*(51.2) = 180 - 102.4 = 77.6°, which is close to 80°, considering the approximation. So that seems consistent.

Alternatively, if we use exact values, since we have x + y = 50°, then angle AFD = 180 - 2(x + y) = 180 - 100 = 80°, which is exact. Therefore, regardless of the exact values of x and y, as long as x + y = 50°, angle AFD would be 80°. Therefore, maybe the answer is 80 degrees.

But wait, I need to make sure that angle FAD is 2x and angle FDA is 2y. Is there a way to confirm that?

Let me think about the diagram again. Since AF is a trisector of angle BAD, which is 3x, then if there are two trisectors, dividing it into three parts. If F is closer to A or D?

Looking at the coordinates of point F: in the Asymptote code, F is at (5, 3). Let me see where that is. Point A is at (0,0), B at (2,4), C at (7,4), D at (7,-2). Point F is at (5,3), which is inside the quadrilateral, closer to the middle. Let me imagine line AF: from A(0,0) to F(5,3). Let me calculate the angle that AF makes with AB and AD.

Vector AF is (5,3). Vector AB is (2,4). Vector AD is (7,-2). Let me compute the angle between AF and AB, and between AF and AD.

First, angle between AF and AB:

Vectors AF = (5,3), AB = (2,4).

Dot product: 5*2 + 3*4 = 10 + 12 = 22

|AF| = sqrt(25 + 9) = sqrt(34) ≈ 5.830

|AB| = sqrt(4 + 16) = sqrt(20) ≈ 4.472

cos(theta) = 22 / (sqrt(34)*sqrt(20)) ≈ 22 / (5.830 * 4.472) ≈ 22 / 26.08 ≈ 0.843

Therefore theta ≈ arccos(0.843) ≈ 32.5°

Similarly, angle between AF and AD:

Vectors AF = (5,3), AD = (7,-2).

Dot product: 5*7 + 3*(-2) = 35 - 6 = 29

|AF| = sqrt(34) ≈ 5.830

|AD| = sqrt(49 + 4) = sqrt(53) ≈ 7.28

cos(theta) = 29 / (5.830 * 7.28) ≈ 29 / 42.43 ≈ 0.683

Therefore theta ≈ arccos(0.683) ≈ 47°

So angle between AF and AB is ≈ 32.5°, angle between AF and AD is ≈ 47°, and the total angle BAD is ≈ 79.5°, which matches. So 32.5 + 47 ≈ 79.5°, which is consistent. Therefore, if angle BAD is trisected into three equal parts, then each trisection angle would be 79.5° / 3 ≈ 26.5°, which matches our previous calculation.

So if angle between AF and AB is 32.5°, which is more than 26.5°, that suggests that AF is not the first trisector (which would be x ≈ 26.5° from AB), but rather the second trisector. Because if the first trisector divides angle BAD into x ≈ 26.5° from AB, then the second trisector would be another x ≈ 26.5° from the first trisector, totaling 53°, leaving the remaining angle to AD as 79.5 - 53 = 26.5°. Wait, but angle between AF and AB is 32.5°, which is larger than x, so that would mean that AF is located between the first trisector and the second trisector? Wait, but there are only two trisectors dividing angle BAD into three equal parts. So if angle between AB and first trisector is x, between first and second trisector is x, between second trisector and AD is x.

Therefore, if AF is located such that the angle between AB and AF is greater than x, then AF must be the second trisector. But according to the calculation, angle between AB and AF is ≈32.5°, which would be 2x, since if x ≈26.5°, then 2x ≈53°, but that's not matching. Wait, wait. Wait, if angle BAD is 3x ≈ 79.5°, then each x ≈26.5°, so from AB, the first trisector is at 26.5°, then the second trisector is at another 26.5°, totaling 53°, then the remaining to AD is 26.5°. Therefore, if AF is creating an angle of 32.5° with AB, which is between the first and second trisectors?

Wait, but according to calculation, angle between AF and AB is ≈32.5°, which would be between the first trisector (26.5°) and second trisector (53°). That suggests that AF is not one of the trisectors, but a different line. But according to the Asymptote code, AF is drawn as a trisector. There's a contradiction here. Maybe my assumption that AF is a trisector is wrong?

Alternatively, maybe angle between AF and AD is 47°, which is roughly 2x (since 2*26.5=53). Close to 47? There's discrepancy here.

Alternatively, maybe my calculation is wrong. Let me recheck.

First, angle between AF and AB: vectors AF=(5,3) and AB=(2,4). The angle between them:

cos(theta) = (AF · AB) / (|AF| |AB|) = (10 + 12)/ (sqrt(34)*sqrt(20)) = 22 / (sqrt(680)) ≈ 22 / 26.07 ≈ 0.844. arccos(0.844) ≈ 32.5°, that's correct. Similarly, angle between AF and AD:

AF · AD = 35 - 6 = 29, |AF|=sqrt(34), |AD|=sqrt(53). So cos(theta)=29 / (sqrt(34*53)) = 29 / sqrt(1802) ≈ 29 / 42.45 ≈ 0.683. arccos(0.683) ≈ 47°, correct.

Therefore, angle between AF and AB is ≈32.5°, angle between AF and AD is ≈47°, and angle BAD is ≈79.5°, which checks out because 32.5 + 47 ≈ 79.5°.

If angle BAD is trisected into three equal parts x, each ≈26.5°, then the first trisector is at ≈26.5° from AB, and second trisector at ≈53° from AB. But AF is at ≈32.5° from AB, which is between the two trisectors. Therefore, AF is not a trisector, which contradicts the problem statement that angles BAD and CDA are trisected as shown. Therefore, my assumption that AF is a trisector must be wrong. Hence, my previous reasoning is flawed.

Alternative approach: Maybe AE and AF are the trisectors. Let me check point E. In the Asymptote code, E is at (4.5 - 0.2, 1 - 0.2) = (4.3, 0.8). So approximately (4.3, 0.8). Let me see where that is in relation to the quadrilateral. Point A is (0,0), B is (2,4), C(7,4), D(7,-2). Point E is near (4.3, 0.8), which is inside the quadrilateral. Let me check the line AE. From A(0,0) to E(4.3, 0.8). Let me calculate the angle between AE and AB, and AE and AD.

Vector AE is (4.3, 0.8). Let's compute angle between AE and AB.

Vector AB is (2,4). The angle between AE and AB:

Dot product: 4.3*2 + 0.8*4 = 8.6 + 3.2 = 11.8

|AE| = sqrt(4.3² + 0.8²) ≈ sqrt(18.49 + 0.64) = sqrt(19.13) ≈ 4.373

|AB| = sqrt(20) ≈ 4.472

cos(theta) = 11.8 / (4.373 * 4.472) ≈ 11.8 / 19.56 ≈ 0.603

Therefore theta ≈ arccos(0.603) ≈ 53°.

Similarly, angle between AE and AD: vector AD is (7, -2). Vector AE is (4.3, 0.8). Dot product: 4.3*7 + 0.8*(-2) = 30.1 - 1.6 = 28.5

|AE| ≈ 4.373, |AD| ≈ 7.28

cos(theta) = 28.5 / (4.373 * 7.28) ≈ 28.5 / 31.83 ≈ 0.895

Therefore theta ≈ arccos(0.895) ≈ 26.5°

Therefore, angle between AE and AB is ≈53°, and angle between AE and AD is ≈26.5°, which suggests that AE is the second trisector of angle BAD, since angle BAD is ≈79.5°, divided into three parts of ≈26.5°, so the first trisector is at 26.5° from AB, the second trisector at 26.5° + 26.5° = 53° from AB, which matches the angle between AE and AB of ≈53°. Therefore, AE is the second trisector, dividing angle BAD into 26.5° and 53°, but since the total angle is 79.5°, that would mean that from AB to AE is 53°, and from AE to AD is 26.5°, which is inconsistent with the trisection. Wait, no: if angle BAD is trisected into three equal parts, then from AB to first trisector is x, first to second trisector is x, and second to AD is x. So total 3x. If x ≈26.5°, then the angle from AB to first trisector is 26.5°, then first to second trisector another 26.5°, totaling 53°, then from second trisector to AD is another 26.5°. Therefore, if AE is the second trisector, then angle between AE and AD is 26.5°, which matches our calculation (≈26.5°). And angle between AE and AB is 53°, which is 2x. So AE is the second trisector, dividing angle BAD into 2x and x.

Similarly, AF is another trisector? But in the diagram, there is also AF drawn. Wait, the Asymptote code draws A--E--D and A--F--D. So AF is another line from A to F. Let's analyze AF.

Point F is at (5,3). Let me calculate angles for AF.

As before, angle between AF and AB is ≈32.5°, and angle between AF and AD is ≈47°. Let me see if that corresponds to any trisection.

If angle BAD is trisected, then possible angles from AB are 26.5°, 53°, and 79.5° (which is AD). If angle between AF and AB is ≈32.5°, which is between the first trisector (26.5°) and second trisector (53°), so AF is not a trisector. However, the problem says that angle BAD is trisected, so there must be lines indicating the trisectors. In the diagram, only AE is drawn as a trisector? Or AF?

Wait, the Asymptote code draws A--E--D and A--F--D. So both AE and AF are drawn from A to E and F, which are connected to D. So maybe AE and AF are both trisectors? But there are two trisectors for angle BAD, so AE and AF could be those two trisectors. Let me check.

If AE is the second trisector (angle from AB to AE is 2x ≈53°), then AF must be the first trisector, with angle from AB to AF being x ≈26.5°. Let me check if AF makes angle ≈26.5° with AB.

Earlier, we calculated angle between AF and AB as ≈32.5°, which is not x ≈26.5°. So this is conflicting.

Wait, unless my estimation is off. Let me check again.

If angle between AF and AB is ≈32.5°, and angle between AE and AB is ≈53°, and angle between AE and AD is ≈26.5°, then if AF is between AB and AE, it might be the first trisector. But according to the angle calculations, it's not matching.

Alternatively, maybe there's a miscalculation. Let me recalculate the angle between AF and AB.

Vector AF is (5,3). Vector AB is (2,4). Their dot product is 5*2 + 3*4 = 10 + 12 = 22. |AF| = sqrt(5² + 3²) = sqrt(34) ≈ 5.830. |AB| = sqrt(2² + 4²) = sqrt(20) ≈ 4.472. So cos(theta) = 22 / (5.830 * 4.472) ≈ 22 / 26.07 ≈ 0.844. The arccos of 0.844 is approximately 32.5°. Let me verify with a calculator. If I take arccos(0.844), it's approximately 32.5°, yes. So that seems correct.

Now, if the first trisector makes an angle of x ≈26.5° with AB, and AF is making 32.5°, which is larger, then AF is not a trisector. This suggests that maybe the diagram's labels are different. Perhaps the x's are not the trisection angles but something else? Or maybe my assumption that angle BAD is 3x is incorrect?

Wait, maybe the trisection is not of the entire angle, but of some other angle? No, the problem says angle BAD and angle CDA are trisected.

Alternatively, maybe the diagram is not drawn to scale, and the coordinates are just for drawing purposes, and the actual angles are different. But the Asymptote code is provided, and the labels of x and y, and the angles 110° and 100°, so maybe I need to use those.

Wait, let's try another approach. Let me use the given angles at B and C, and the fact that quadrilateral ABCD has angles at B = 110°, at C = 100°, and we have angles at A and D to find.

From earlier, we have angle A + angle D = 360 - 110 - 100 = 150°. So angle A + angle D = 150°. If angle BAD is 3x and angle CDA is 3y, then 3x + 3y = 150 => x + y = 50°, as before.

Now, if I can relate angle AFD to x and y, I need to find relations in triangle AFD or other triangles.

Let me consider triangle AFD. To find angle AFD, I need to know two other angles. Let's consider angles at A and D.

Angle at A in triangle AFD is angle FAD. Angle at D in triangle AFD is angle FDA.

If I can express these angles in terms of x and y, then I can find angle AFD.

Assuming that AF is a trisector of angle BAD, then angle FAD is either x or 2x. Similarly, if FD is a trisector of angle CDA, then angle FDA is y or 2y.

But how to determine which?

Let me consider the Asymptote code labels. There are three x's and three y's. The x's are at positions that might correspond to the three trisection angles of angle BAD. Similarly for y. If angle BAD is trisected into three angles, each labeled x, then the sum of those three angles is 3x = angle BAD. Similarly for angle CDA.

If there are three x's, perhaps each of the trisection angles is labeled x. For example, between AB and first trisector: x, between first and second trisectors: x, between second and AD: x. Similarly for y.

Now, in the diagram, there are two lines from A: AE and AF. If these are the two trisectors, then one of them is the first trisector and the other is the second trisector.

If I assume that AE is the second trisector (as we saw earlier, it makes angle ≈53° with AB, which is 2x where x ≈26.5°), and AF is the first trisector, making angle x ≈26.5° with AB. But earlier calculation suggests that AF makes angle ≈32.5° with AB, which is not x ≈26.5°. However, maybe there's a calculation error?

Wait, let me recalculate the angle between AF and AB. Maybe I made a mistake in the coordinates of F?

Wait, in the Asymptote code, F is at (5, 3). Let me confirm that. The Asymptote code says:

pair E, F;
E = (4.5-.2,1-.2); F = (5, 3);

So E is at (4.5 - 0.2, 1 - 0.2) = (4.3, 0.8), and F is at (5,3). So F is at (5, 3). Let me check the angle between AF and AB again.

Vector AF is from A(0,0) to F(5,3): (5,3). Vector AB is from A(0,0) to B(2,4): (2,4).

The angle between (5,3) and (2,4). Let me calculate this angle using the tangent formula.

The slope of AF is 3/5 = 0.6, and the slope of AB is 4/2 = 2. The angle between them can be calculated as:

tan(theta) = |(m2 - m1)/(1 + m1*m2)| = |(2 - 0.6)/(1 + (2)(0.6))| = |1.4 / 2.2| ≈ 0.6364

Therefore theta ≈ arctan(0.6364) ≈ 32.5°, which confirms the previous result. So angle between AF and AB is ≈32.5°, which is greater than x ≈26.5°, suggesting AF is not the first trisector.

But if AE is the second trisector, then angle between AE and AB is ≈53°, which is 2x, and angle between AE and AD is x ≈26.5°, as calculated earlier. So if AE is the second trisector, then the first trisector would be somewhere between AB and AE, making x ≈26.5° with AB. Let's call this first trisector line AG, where G is some point. If the first trisector is AG, then angle between AB and AG is x ≈26.5°, and between AG and AE is x ≈26.5°, and between AE and AD is x ≈26.5°.

But in the diagram, there is no line AG, only AE and AF. So perhaps AF is a different line? Maybe it's a trisector of angle CDA?

Wait, the problem says angle BAD and angle CDA are trisected. So both angles are trisected, with two trisectors each. In the diagram, from A, there are two lines: AE and AF. From D, there might be two trisectors, but only one is shown, FD. So FD is a trisector of angle CDA.

If FD is a trisector of angle CDA, then let's calculate the angle between DC and DF, and between DF and DA.

Point F is at (5,3). Let me calculate the angle between DF and DC, and between DF and DA.

Vector DF is from D(7,-2) to F(5,3): (5-7, 3 - (-2)) = (-2, 5).

Vector DC is from D(7,-2) to C(7,4): (0,6).

Vector DA is from D(7,-2) to A(0,0): (-7, 2).

First, let's find the angle between DF and DC.

Vector DF = (-2,5), vector DC = (0,6).

Dot product: (-2)(0) + (5)(6) = 30.

|DF| = sqrt(4 + 25) = sqrt(29) ≈ 5.385

|DC| = 6

cos(theta) = 30 / (5.385 * 6) ≈ 30 / 32.31 ≈ 0.9287

So theta ≈ arccos(0.9287) ≈ 21.8°

Similarly, angle between DF and DA: vector DF = (-2,5), vector DA = (-7,2).

Dot product: (-2)(-7) + (5)(2) = 14 + 10 = 24.

|DF| ≈ 5.385, |DA| ≈ 7.28

cos(theta) = 24 / (5.385 * 7.28) ≈ 24 / 39.1 ≈ 0.614

Therefore theta ≈ arccos(0.614) ≈ 52°.

So angle between DF and DC is ≈21.8°, and between DF and DA is ≈52°. The total angle CDA is ≈74°, so 21.8 + 52 ≈73.8°, which matches. Therefore, if FD is a trisector of angle CDA, then angle between DC and DF is y, and between DF and DA is 2y. So y ≈21.8°, and 2y ≈43.6°, but we calculated 52°, which is inconsistent. Alternatively, if FD is dividing angle CDA into y and 2y, then 3y = 74°, so y ≈24.7°, but our calculation gave angle between DC and DF as ≈21.8°, and between DF and DA as ≈52°, which is not a multiple of y. This inconsistency suggests that FD is not a trisector, or my calculations are wrong.

Wait, but if angle CDA is trisected, then it should be divided into three equal parts. If angle CDA is 74°, then each trisection angle is ≈24.7°. Let me check if FD divides it into 24.7° and 49.4°, but our calculation showed 21.8° and 52°, which doesn't align. So something is wrong here.

Alternatively, perhaps the labels in the Asymptote code are not corresponding to the trisection angles but to other parts. Given the confusion, perhaps I should use the earlier relation x + y = 50° and assume that in triangle AFD, angle AFD = 180 - 2x - 2y = 80°, regardless of the exact positions, since x + y = 50°.

But why would angle AFD be 180 - 2x - 2y? If in triangle AFD, angle at A is 2x and angle at D is 2y, then yes. But how do I know those are the correct angles?

Alternatively, let me think about the problem differently. Since angle BAD is trisected, let me call the trisectors as follows: from A, draw two lines, dividing angle BAD into three equal parts. Let's call the first trisector as line AM, making angle x with AB, and second trisector as line AN, making 2x with AB. Similarly, for angle CDA, trisectors: from D, draw two lines, dividing angle CDA into three equal parts. Let's call the first trisector as line DP, making angle y with DC, and second trisector as line DQ, making 2y with DC.

Now, if in the diagram, the lines drawn are AM and AN from A, and DP and DQ from D. But in the Asymptote code, only lines from A to E and A to F, and from F to D. So perhaps F is the intersection of two trisectors? Or maybe F is on a trisector of angle CDA?

If F is on the trisector of angle CDA, then DF would be that trisector, and AF would be some line. But without more information, it's hard to say.

Alternatively, using the given labels in the diagram, with x and y, and given that x + y = 50°, and looking for angle AFD, which is likely 80° as previously calculated, and since this is a common angle in such trisection problems, I might conclude that the answer is 80°.

However, to confirm, let me think about the diagram's angles. Let me assume that angle AFD is 80°, and see if it makes sense with the other angles.

If x + y = 50°, and angle AFD = 180 - 2(x + y) = 80°, this is a solid relation, assuming that angle FAD = 2x and angle FDA = 2y.

Alternatively, if angle FAD = x and angle FDA = y, then angle AFD = 180 - x - y = 130°, but this would require that AF is a trisector of angle BAD and FD is a trisector of angle CDA, but based on the earlier calculations, it seems that AF is not a trisector. However, if this were the case, then angle AFD = 130°, but I need to check if this is possible.

Wait, let me think about the given labels. The Asymptote code labels three x's and three y's. If angle BAD is trisected into three x's, then each trisection angle is x, so x = angle BAD / 3. Similarly for y. If the three x's are the trisection angles, then in triangle AFD, if angle FAD is x and angle FDA is y, then angle AFD = 180 - x - y. But since x + y = 50°, then angle AFD = 130°. But why would angle FAD be x and angle FDA be y?

If AF is the first trisector of angle BAD, then angle FAD = 2x (since from AF to AD is x, and from AB to AF is x, total angle BAD = 3x). Wait, if AF is the first trisector, then angle between AB and AF is x, and between AF and AD is 2x. So angle FAD is 2x. Similarly, if FD is the first trisector of angle CDA, then angle between DC and FD is y, and between FD and DA is 2y. So angle FDA is 2y. Hence, angle AFD = 180 - 2x - 2y = 80°, as before.

But if AF is the second trisector, then angle FAD = x, and if FD is the second trisector, angle FDA = y, then angle AFD = 180 - x - y = 130°. But without knowing which trisectors AF and FD are, it's hard to say.

However, given that there are three x's and three y's in the diagram, it's more likely that each trisection angle is labeled, and therefore, angle FAD and angle FDA are multiples of x and y. Since x + y = 50°, and the only way to get a unique answer is if angle AFD is 80°, which is 180 - 2(x + y), I think this is the intended solution.

Therefore, I will go with angle AFD = 80°, so the degree measure is 80 degrees.

**Final Answer**
\boxed{80}

To determine the measure of angle $ \angle AFD $ in quadrilateral $ ABCD $ with trisected angles $ \angle BAD $ and $ \angle CDA $, we proceed step-by-step using geometric reasoning and algebraic relations.

---

### **1. Understanding the Geometry**

We are given that:

- Quadrilateral $ ABCD $ includes angles $ \angle BAD $ and $ \angle CDA $, which are each **trisected**.
- Points $ E $ and $ F $ lie on the trisectors from $ A $ and $ D $, respectively.
- Diagram indicates that lines $ AE $ and $ AF $ originate from vertex $ A $, while $ DF $ extends from $ D $.
- Labeling in the diagram suggests that the trisectors split $ \angle BAD $ into three equal parts, each of measure $ x $, and $ \angle CDA $ into three equal parts, each of measure $ y $.

From this, we can assign:

- $ \angle BAD = 3x $
- $ \angle CDA = 3y $
- Angles at $ B $ and $ C $ are given as $ 110^\circ $ and $ 100^\circ $, respectively.

---

### **2. Using the Angle Sum of a Quadrilateral**

The sum of the interior angles in any quadrilateral is:

$$
\text{Sum of angles} = 360^\circ
$$

Thus:

$$
\angle BAD + \angle CDA + \angle ABC + \angle BCD = 360^\circ
$$

$$
3x + 3y + 110^\circ + 100^\circ = 360^\circ
\Rightarrow 3x + 3y = 150^\circ
\Rightarrow x + y = 50^\circ
$$

This is our key relationship:
$$
x + y = 50^\circ
$$

---

### **3. Analyzing Triangle $ AFD $**

We are to find $ \angle AFD $, the angle at vertex $ F $ in triangle $ AFD $. Let’s express this angle using known quantities.

In triangle $ AFD $, the three angles are:

- $ \angle FAD $: the angle at vertex $ A $, formed by the trisection of $ \angle BAD $. Depending on which trisector corresponds to ray $ AF $, $ \angle FAD = 2x $ (if $ AF $ is the second trisector from $ A $).
- $ \angle FDA $: the angle at vertex $ D $, formed by the trisection of $ \angle CDA $. Similarly, if $ FD $ is the second trisector from $ D $, $ \angle FDA = 2y $.

Thus, the angle at $ F $ is:

$$
\angle AFD = 180^\circ - \angle FAD - \angle FDA = 180^\circ - 2x - 2y
$$

Now, substituting $ x + y = 50^\circ $:

$$
\angle AFD = 180^\circ - 2(x + y) = 180^\circ - 2(50^\circ) = 180^\circ - 100^\circ = \boxed{80^\circ}
$$

---

### **4. Final Result**

Thus, the degree measure of angle $ \angle AFD $ is:

$$
\boxed{80}
$$

Solution: Triangle $AFD$ must have a total angle measure of $180^\circ$. We know that the other two angles have measures of $2x$ and $2y$, so angle $AFD$ must have a measure of $180-2x-2y=180-(2x+2y)$ degrees. We now look at quadrilateral $ABCD$, whose interior angle measures must sum to $360^\circ$. Therefore, we have that $110^\circ +100^\circ +3y+3x=360^\circ$, so $3x+3y=150^\circ$. We want to find $2x+2y$, so we multiply both sides of the equation by $2/3$ to get that $2x+2y=100^\circ$. We can now substitute in $100^\circ$ for $2x+2y$ to find that the measure of angle $AFD$ is $180-(2x+2y)=180-100=\boxed{80}$ degrees.