Explanation:
Okay, so I need to find the measure of angle AFD in this quadrilateral ABCD. The problem says that angles BAD and CDA are trisected, and there's an Asymptote figure provided. Let me try to visualize this first.

First, let me recall what trisecting an angle means. Trisecting an angle means dividing it into three equal parts. So, if angle BAD is trisected, there will be two lines from point A that divide angle BAD into three equal angles. Similarly, angle CDA is trisected, so there will be two lines from point D that divide angle CDA into three equal angles. The figure shows points E and F, which are probably the trisection points.

Looking at the Asymptote code, points E and F are defined. E is near (4.5 - 0.2, 1 - 0.2), which is approximately (4.3, 0.8), and F is at (5, 3). The labels for angles x and y are placed in different positions. There's also a label for 110 degrees near (2.5, 3.5) and 100 degrees near (6.5 - 0.2, 3.5).

Wait, maybe I should try to figure out the measures of angles BAD and CDA first? Because if they are trisected, then knowing those angles would help me find the trisection angles.

But how do I find angles BAD and CDA? The Asymptote code gives coordinates for points A, B, C, D. Let me try to use coordinates to calculate the angles.

Coordinates:
A = (0,0)
B = (2,4)
C = (7,4)
D = (7,-2)

So quadrilateral ABCD has coordinates A(0,0), B(2,4), C(7,4), D(7,-2). Let me sketch this mentally. From A(0,0) to B(2,4) to C(7,4) to D(7,-2) back to A.

First, let me compute angle BAD. That's the angle at point A between BA and DA. Similarly, angle CDA is the angle at point D between CD and DA.

To compute angle BAD, I can find vectors AB and AD and then compute the angle between them. Similarly for angle CDA.

First, vector AB is from A(0,0) to B(2,4): so vector AB is (2,4). Vector AD is from A(0,0) to D(7,-2): so vector AD is (7,-2).

The angle between AB and AD is angle BAD. Let me compute that using the dot product formula.

The formula for the angle θ between vectors u and v is:

cosθ = (u · v)/(|u| |v|)

So, compute AB · AD = (2)(7) + (4)(-2) = 14 - 8 = 6

|AB| = sqrt(2² + 4²) = sqrt(4 + 16) = sqrt(20) = 2*sqrt(5)

|AD| = sqrt(7² + (-2)²) = sqrt(49 + 4) = sqrt(53)

Therefore, cosθ = 6 / (2√5 * √53) = 6 / (2√265) = 3 / √265 ≈ 3 / 16.2788 ≈ 0.1844

Then θ ≈ arccos(0.1844) ≈ 80 degrees? Wait, let me check with calculator.

Wait, arccos(0.1844). Let me recall that cos(80°) ≈ 0.1736, cos(75°) ≈ 0.2588. So 0.1844 is between 75° and 80°, closer to 80°. Let me compute it more accurately.

Let me use a calculator. Let me compute arccos(3 / sqrt(265)).

First, sqrt(265) ≈ 16.2788, so 3 / 16.2788 ≈ 0.1844.

Now, arccos(0.1844). Let me convert this to degrees.

Using a calculator, arccos(0.1844) ≈ 80 degrees? Let me check with calculator steps.

Alternatively, maybe I can compute it using inverse cosine. But since I don't have a calculator here, maybe I can use approximate values.

Alternatively, maybe there's another way. Wait, but maybe I can find angle BAD using coordinates. Alternatively, maybe the problem gives some angles? Wait, the Asymptote code labels 110° and 100°. Let me check the labels.

The Asymptote code labels "110°" at (2.5, 3.5) and "100°" at (6.5 - 0.2, 3.5). Let me see where those points are.

Point (2.5, 3.5) is near the middle of AB and BC? Wait, AB is from (0,0) to (2,4). The midpoint of AB is (1,2). But (2.5, 3.5) is near point B(2,4) and C(7,4). Wait, maybe that's angle ABC? Or angle at B? Wait, the label is near (2.5, 3.5), which is above AB. Maybe that's angle ABC? Wait, but angle ABC is at point B. Let me check coordinates.

Alternatively, maybe those labels are for angles that are parts of the trisection? Wait, the problem says angles BAD and CDA are trisected. The Asymptote code labels "x" and "y" in different positions. Maybe those are the trisection angles?

Wait, the problem says "angle BAD and angle CDA are trisected as shown." So maybe the diagram shows trisection lines, and the labels x and y are the measures of those trisection angles? But the problem also labels 110° and 100°, which might be angles at B and C?

Alternatively, maybe the 110° and 100° are angles that are part of the trisection? Wait, the Asymptote code labels "110°" near (2.5, 3.5) which is near the middle of AB and BC. Maybe that's angle ABC? Let me check coordinates.

Point B is at (2,4). The angle at B is between AB and BC. Let me compute angle ABC.

Vector BA is from B to A: (-2, -4). Vector BC is from B to C: (5, 0).

So angle between BA and BC. Let me compute that.

Dot product: (-2)(5) + (-4)(0) = -10 + 0 = -10

|BA| = sqrt(4 + 16) = sqrt(20) = 2√5

|BC| = sqrt(25 + 0) = 5

So cosθ = (-10)/(2√5 * 5) = (-10)/(10√5) = -1/√5 ≈ -0.4472

Therefore θ ≈ 116.565 degrees. Wait, but the label says 110°. Hmm, maybe my calculation is wrong? Wait, maybe I need to check again.

Wait, angle ABC is at point B. The vectors are BA and BC. Wait, BA is from B to A, which is (-2, -4). BC is from B to C, which is (5, 0). So the angle between BA and BC is indeed calculated as above.

But according to calculation, it's arccos(-1/√5) ≈ 116.565°, but the diagram labels it as 110°. Maybe there's a discrepancy here? Or maybe my coordinates are wrong? Wait, the Asymptote code defines A(0,0), B(2,4), C(7,4), D(7,-2). So coordinates are correct.

Alternatively, maybe the diagram is not to scale? Because the problem says "as shown," but since I don't have the image, maybe the labels 110° and 100° are given for angles at B and C? Wait, the label "110°" is at (2.5, 3.5), which is near the middle of AB and BC. Maybe that's angle ABC? But according to calculation, it's about 116.5°, not 110°. Maybe the diagram is approximate?

Alternatively, maybe the 110° and 100° are angles that are parts of the trisection? Wait, the problem says angles BAD and CDA are trisected. 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 are those trisection angles?

Looking at the Asymptote code, there are labels for x in positions (1, 1.5), (2, 1), (2, -1). Maybe those are the trisection angles? Similarly, labels for y near (5.5, 0.5), (6.5, 0), (5, -1.5).

But without seeing the diagram, it's a bit challenging. However, maybe I can proceed with the coordinates.

Alternatively, maybe the problem is from a known source? Wait, the Asymptote code is given, so maybe I can reconstruct the diagram.

Let me try to think again. The quadrilateral ABCD with coordinates A(0,0), B(2,4), C(7,4), D(7,-2). So AB is from (0,0) to (2,4), BC is from (2,4) to (7,4), CD is from (7,4) to (7,-2), and DA is from (7,-2) back to (0,0).

So angle BAD is at point A between BA and DA. As calculated earlier, angle BAD ≈ 80 degrees? Wait, earlier calculation gave angle BAD ≈ arccos(3 / sqrt(265)) ≈ 80 degrees? Let me check with approximate values.

Alternatively, maybe I can compute the slopes of AB and AD to find the angle between them.

Slope of AB: from A(0,0) to B(2,4): rise over run is 4/2 = 2. So slope is 2.

Slope of AD: from A(0,0) to D(7,-2): rise over run is -2/7 ≈ -0.2857.

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

tanθ = |(m2 - m1)/(1 + m1*m2)|

So tanθ = |(-2/7 - 2)/(1 + (2)(-2/7))| = |(-2/7 - 14/7)/(1 - 4/7)| = |(-16/7)/(3/7)| = | -16/3 | = 16/3 ≈ 5.333...

Therefore θ = arctan(16/3) ≈ arctan(5.333). Let me recall that tan(79°) ≈ 5.145, tan(80°) ≈ 5.671. So 16/3 ≈ 5.333 is between 79° and 80°, closer to 80°. Let me compute it more accurately.

Let me use linear approximation. Let’s say tan(79°) ≈ 5.145, tan(80°) ≈ 5.671. The difference between tan(80°) and tan(79°) is approximately 0.526 per degree. Our value is 5.333, which is 5.333 - 5.145 = 0.188 above tan(79°). So fraction is 0.188 / 0.526 ≈ 0.357. So angle ≈ 79° + 0.357° ≈ 79.357°, approximately 79.4 degrees.

So angle BAD ≈ 79.4 degrees. If this is trisected, then each trisection angle would be approximately 79.4 / 3 ≈ 26.5 degrees.

Similarly, angle CDA is at point D between CD and DA. Let me compute that angle.

Point D is at (7,-2). CD is from D(7,-2) to C(7,4), which is straight up along the vertical line x=7. DA is from D(7,-2) to A(0,0).

So vector DC is (0,6) (from D to C), and vector DA is (-7, 2) (from D to A).

So angle between DC and DA is angle CDA. Let me compute that.

Vectors DC = (0,6) and DA = (-7, 2).

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

|DC| = sqrt(0² + 6²) = 6

|DA| = sqrt((-7)^2 + 2^2) = sqrt(49 + 4) = sqrt(53) ≈ 7.28

Therefore cosθ = 12 / (6 * sqrt(53)) = 2 / sqrt(53) ≈ 2 / 7.28 ≈ 0.275

Therefore θ ≈ arccos(0.275) ≈ 73.9 degrees.

So angle CDA ≈ 73.9 degrees. If trisected, each trisection angle would be ≈ 73.9 / 3 ≈ 24.6 degrees.

But the problem mentions that angles BAD and CDA are trisected, and there are labels for x and y. However, in the Asymptote code, there are labels for x and y in different positions. Maybe those x and y are the trisection angles?

But the problem also labels 110° and 100°, which might be angles at B and C? Wait, earlier calculation for angle ABC was approximately 116.5°, but the diagram labels it as 110°, so maybe there's inconsistency.

Alternatively, maybe the given angles of 110° and 100° are the measures of angles BAD and CDA? Wait, if angle BAD is 110°, then trisected into three angles of 110/3 ≈ 36.66°, and angle CDA is 100°, trisected into 100/3 ≈ 33.33°. But earlier calculations with coordinates gave angle BAD ≈ 79.4° and angle CDA ≈ 73.9°, which don't match 110° and 100°.

This suggests that maybe the diagram is not drawn to scale, or perhaps the labels 110° and 100° are not angles BAD and CDA but something else.

Alternatively, maybe the 110° and 100° are angles formed by the trisection lines?

Wait, the Asymptote code draws A--E--D and A--F--D. So E and F are points on AD? Wait, no. Wait, E is defined as (4.5 - 0.2, 1 - 0.2) which is approximately (4.3, 0.8). Let me check if that's on any line.

Looking at the coordinates, point E is near (4.3, 0.8). Let me see if that's on AB or AD or BC or CD?

Point A is (0,0), B is (2,4), C is (7,4), D is (7,-2).

If I consider the line from A(0,0) to E(4.3, 0.8), that would be a line from A to somewhere inside the quadrilateral. Similarly, line from A to F(5,3).

But the Asymptote code draws A--E--D and A--F--D. So E and F are points such that AE and AF are trisectors of angle BAD, and DE and DF are trisectors of angle CDA?

Wait, the problem says angles BAD and CDA are trisected. So angle BAD is trisected, meaning there are two lines from A dividing angle BAD into three equal parts. Similarly, angle CDA is trisected, meaning two lines from D dividing angle CDA into three equal parts.

In the diagram, points E and F are drawn. The Asymptote code draws A--E--D and A--F--D. So E and F are points on AD? Wait, no. If you draw A--E--D, then E is a point such that AE and ED are parts of the trisection? Wait, no. If angle BAD is trisected, then the trisectors would be lines from A, not from D. Similarly, angle CDA is trisected, so trisectors from D.

Wait, maybe E is on AB and F is on DC? Or somewhere else?

Alternatively, maybe E and F are points where the trisectors intersect some sides?

This is getting confusing without seeing the diagram. Maybe I need to think differently.

The problem asks for the measure of angle AFD. Points A, F, D. So angle at F between A and D.

Given that F is a trisection point of angle CDA? Or angle BAD?

Wait, the Asymptote code defines E and F as points. Let me check their coordinates.

E is defined as (4.5 - 0.2, 1 - 0.2) which is (4.3, 0.8). Let me see if that's on any line.

Looking at the coordinates, point B is (2,4), C is (7,4). The line BC is horizontal from (2,4) to (7,4). The point E is at (4.3, 0.8), which is below BC.

Alternatively, maybe E is on BD? Let me check.

Point B is (2,4), D is (7,-2). The line BD can be parametrized. Let me see if (4.3, 0.8) is on BD.

Parametric equations for BD: starting at B(2,4), direction vector (5, -6). So parametric equations: x = 2 + 5t, y = 4 - 6t.

Let me see if there's a t such that x = 4.3 and y = 0.8.

From x: 2 + 5t = 4.3 => 5t = 2.3 => t = 0.46

Then y = 4 - 6*(0.46) = 4 - 2.76 = 1.24. But the y-coordinate of E is 0.8, not 1.24. So not on BD.

Alternatively, maybe E is on AB? Let me check. AB is from (0,0) to (2,4). Let me parametrize AB: x = 2s, y = 4s, where s ∈ [0,1].

If E is (4.3, 0.8), then 2s = 4.3 => s = 2.15, which is outside the range [0,1]. So not on AB.

Similarly, check if E is on AD. AD is from (0,0) to (7,-2). Parametric equations: x = 7t, y = -2t.

If E is (4.3, 0.8), then 7t = 4.3 => t ≈ 0.614, then y = -2*(0.614) ≈ -1.228, but E's y-coordinate is 0.8. Not matching.

So E is not on AD. Maybe E is on BC? BC is from (2,4) to (7,4). So y-coordinate is 4. But E has y=0.8, so no.

Alternatively, maybe E is inside the quadrilateral?

Similarly, point F is at (5,3). Let me check if that's on any line.

Looking at coordinates, point F(5,3). Let me check if it's on AB, BC, CD, or DA.

AB: from (0,0) to (2,4). Parametric equations: x=2s, y=4s. If x=5, then s=2.5, which is outside.

BC: horizontal line y=4. F has y=3, so no.

CD: vertical line x=7 from (7,4) to (7,-2). F has x=5, so no.

DA: from (7,-2) to (0,0). Parametric equations: x=7-7t, y=-2 + 2t. If x=5, then 7 -7t =5 => t=2/7. Then y = -2 + 2*(2/7) = -2 + 4/7 ≈ -1.428, which is not 3.

So F is not on any side. Therefore, E and F are points inside the quadrilateral.

Given that, and the Asymptote code draws A--E--D and A--F--D. So AE and AF are lines from A to E and F, and ED and FD are lines from E and F to D.

But the problem says angles BAD and CDA are trisected. So angle BAD is trisected by AE and AF? Or maybe AE and another line?

Wait, angle BAD is at point A. If it's trisected, there should be two lines from A dividing angle BAD into three equal angles. Similarly, angle CDA is at point D, so two lines from D dividing angle CDA into three equal angles.

In the diagram, there are two lines from A: AE and AF. Maybe those are the trisectors of angle BAD? Similarly, maybe there are two lines from D, but in the diagram, only FD is drawn? Wait, the Asymptote code draws A--E--D and A--F--D. So from A, two lines AE and AF going to E and F, and then from E and F to D.

But if angle BAD is trisected, then AE and AF would be the trisectors. Similarly, if angle CDA is trisected, then maybe FD is one of the trisectors?

But the problem says both angles BAD and CDA are trisected. So maybe there are two trisectors for each angle.

But since the diagram shows only two lines from A (AE and AF) and one line from D (FD), maybe FD is a trisector of angle CDA?

Alternatively, maybe there are two trisectors from D, but only FD is shown?

This is getting too ambiguous without the diagram. Maybe I need to use the given labels.

The Asymptote code labels "x" in three places: (1, 1.5), (2, 1), (2, -1). Maybe those are the trisection angles of angle BAD? Similarly, labels "y" in three places: (5.5, 0.5), (6.5, 0), (5, -1.5). Maybe those are trisection angles of angle CDA?

If that's the case, then angle BAD is divided into three angles of measure x, and angle CDA is divided into three angles of measure y.

Additionally, there are labels "110°" and "100°". Maybe those are angles at B and C?

Alternatively, maybe the 110° and 100° are the measures of angles BAD and CDA?

Wait, if angle BAD is 110°, then each trisection angle is 110/3 ≈ 36.66°, and angle CDA is 100°, each trisection angle is 100/3 ≈ 33.33°.

But earlier calculations with coordinates gave angle BAD ≈ 79.4° and angle CDA ≈ 73.9°, which don't match.

Alternatively, maybe the 110° and 100° are angles formed by the trisectors?

Alternatively, maybe the problem is from a specific source where the diagram is standard, and the angles BAD and CDA are 110° and 100°, which are trisected.

Wait, the problem says "angle BAD and angle CDA are trisected as shown." So maybe the diagram shows that angle BAD is trisected into three angles, and angle CDA is trisected into three angles, and the labels 110° and 100° are for those angles?

But if angle BAD is trisected, then the total angle would be 3x, and angle CDA would be 3y.

But the problem doesn't give specific values for x or y, but labels them as x and y. However, there are also labels for 110° and 100°, which might be angles at B and C?

Alternatively, maybe the 110° and 100° are the measures of angles BAD and CDA?

If angle BAD is 110°, then each trisection angle is 110/3 ≈ 36.66°, and angle CDA is 100°, each trisection angle is 100/3 ≈ 33.33°.

But then, how does that help me find angle AFD?

Alternatively, maybe the diagram is such that angle BAD is 110°, angle CDA is 100°, and they are trisected, so we can find the measure of angle AFD using those trisection angles.

Let me assume that angle BAD = 110°, trisected into three angles of 110/3 ≈ 36.66°, and angle CDA = 100°, trisected into three angles of 100/3 ≈ 33.33°.

Then, points E and F are the trisection points.

But how does that help me find angle AFD?

Let me think about triangle AFD. To find angle AFD, I need to know some information about triangle AFD. Maybe I can find the measures of angles at A and D in triangle AFD, then use the fact that the sum of angles in a triangle is 180°.

But angle at A in triangle AFD is part of angle BAD. Similarly, angle at D in triangle AFD is part of angle CDA.

Assuming that angle BAD is trisected, then if AF is one of the trisectors, then angle FAD would be one-third of angle BAD. Similarly, if FD is one of the trisectors of angle CDA, then angle FDA would be one-third of angle CDA.

But I need to figure out which trisectors AF and FD are.

Looking back at the Asymptote code, the label "x" is near (1, 1.5), (2, 1), (2, -1). Maybe those are the trisection angles of angle BAD. Similarly, labels "y" near (5.5, 0.5), (6.5, 0), (5, -1.5) are trisection angles of angle CDA.

If angle BAD is trisected into three angles of measure x, then angle BAD = 3x. Similarly, angle CDA = 3y.

Additionally, there are labels for 110° and 100°, which might be angles at B and C.

But without knowing the exact configuration, it's challenging. However, maybe I can use the fact that in quadrilateral ABCD, the sum of all interior angles is 360°.

Wait, quadrilateral ABCD has angles at A, B, C, D.

If I can find angles at A, B, C, D, then maybe I can find relations.

Earlier, I calculated angle BAD ≈ 79.4°, angle ABC ≈ 116.5°, angle BCD, and angle CDA ≈ 73.9°. Let me compute angle BCD.

Point C is at (7,4). Angle BCD is at point C between BC and CD.

Vector CB is from C to B: (-5, 0). Vector CD is from C to D: (0, -6).

So angle between CB and CD.

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

Wait, that means the angle is 90 degrees? Because the dot product is zero.

So angle BCD is 90 degrees.

Therefore, sum of angles in quadrilateral ABCD: angle A + angle B + angle C + angle D = 79.4 + 116.5 + 90 + 73.9 ≈ 359.8°, which is roughly 360°, considering rounding errors.

So angle BAD ≈ 79.4°, angle ABC ≈ 116.5°, angle BCD = 90°, angle CDA ≈ 73.9°.

But the problem labels 110° and 100°, which might be approximate values for angles at B and C?

If angle ABC is labeled as 110°, and angle BCD is labeled as 100°, but according to calculation, angle ABC is ≈ 116.5°, angle BCD is 90°, which doesn't match.

Alternatively, maybe the labels 110° and 100° are for angles BAD and CDA?

If angle BAD is 110°, angle CDA is 100°, then angle ABC and angle BCD can be calculated.

Let me try that.

Assume angle BAD = 110°, angle CDA = 100°, angle BCD = 90°, then angle ABC = 360 - 110 - 100 - 90 = 60°.

But according to coordinates, angle ABC was ≈ 116.5°, which conflicts.

Alternatively, maybe the diagram is different.

Alternatively, maybe the given angles of 110° and 100° are the measures of angles formed by the trisectors?

This is getting too confusing. Maybe I need to approach this problem using the trisection angles and the given labels.

Let me consider that angle BAD is trisected into three angles of measure x, and angle CDA is trisected into three angles of measure y.

From the Asymptote code, there are labels for x and y. Also, there are labels for 110° and 100°. Maybe those are angles that are formed by the trisectors?

Alternatively, maybe the 110° and 100° are angles at B and C, and using that, we can find x and y.

Let me try that.

Assume angle ABC = 110°, angle BCD = 100°, and angle BAD = 3x, angle CDA = 3y.

Then, in quadrilateral ABCD, sum of angles is 360°, so:

angle BAD + angle ABC + angle BCD + angle CDA = 3x + 110 + 100 + 3y = 360

So 3x + 3y = 360 - 110 - 100 = 150

Therefore, x + y = 50°

So x + y = 50°.

Now, need more equations to find x and y.

Alternatively, maybe using triangle AFD.

But how?

Alternatively, consider triangle AFD. To find angle AFD, we need to know some relations.

Assuming that AF is a trisector of angle BAD, so angle FAD = x, and FD is a trisector of angle CDA, so angle FDA = y.

Then, in triangle AFD, angles at A and D are x and y, so angle AFD = 180° - x - y = 180° - (x + y) = 180° - 50° = 130°.

Wait, that seems straightforward! If angle FAD = x and angle FDA = y, then angle AFD = 180 - x - y = 130°.

But is this valid?

Wait, but why would angle FAD = x and angle FDA = y?

If AF is a trisector of angle BAD, then angle FAD would be one-third of angle BAD, which is x. Similarly, if FD is a trisector of angle CDA, then angle FDA would be one-third of angle CDA, which is y.

But is FD a trisector of angle CDA?

In the diagram, FD is drawn from F to D. If angle CDA is trisected, then there should be two lines from D dividing angle CDA into three equal parts. If FD is one of those trisectors, then angle FDA would be y.

But how do we know that FD is a trisector?

Alternatively, maybe AF is a trisector of angle BAD, and FD is a trisector of angle CDA.

But without seeing the diagram, it's hard to be certain. However, given that the problem labels angles x and y, and given that angle BAD = 3x and angle CDA = 3y, and from the quadrilateral angle sum we found x + y = 50°, then angle AFD = 180 - x - y = 130°.

But the problem might have different configuration.

Wait, but if angle AFD is 130°, is that the answer?

But let me check with the coordinates.

Earlier, I calculated angle BAD ≈ 79.4°, angle CDA ≈ 73.9°, so x ≈ 26.5°, y ≈ 24.6°, so x + y ≈ 51.1°, so angle AFD ≈ 180 - 51.1 ≈ 128.9°, which is roughly 130°. So maybe the answer is 130°?

But the problem might have exact values. If angle BAD and CDA are trisected, and with the given labels of 110° and 100°, which might be angles at B and C, leading to x + y = 50°, then angle AFD = 130°.

Alternatively, maybe the labels 110° and 100° are for angles BAD and CDA?

If angle BAD = 110°, angle CDA = 100°, then x = 110/3 ≈ 36.66°, y = 100/3 ≈ 33.33°, so x + y ≈ 70°, angle AFD = 180 - 70 = 110°. But this contradicts previous calculation.

But how do I know which is correct?

Wait, the Asymptote code labels "110°" near (2.5, 3.5) and "100°" near (6.5 - 0.2, 3.5). Let me think about where those points are.

Point (2.5, 3.5) is near the middle of AB and BC? If AB is from (0,0) to (2,4), and BC is from (2,4) to (7,4). The point (2.5, 3.5) is between AB and BC. Maybe that's angle ABC? But earlier calculation said angle ABC ≈ 116.5°, but labeled as 110°.

Similarly, the label "100°" is near (6.5 - 0.2, 3.5) which is (6.3, 3.5). That's near point C(7,4). Maybe that's angle BCD? Earlier calculation said angle BCD is 90°, but labeled as 100°.

This inconsistency suggests that the diagram is not to scale, or the labels are for different angles.

Alternatively, maybe the 110° and 100° are the measures of angles formed by the trisectors?

For example, if angle BAD is trisected into three angles, and one of them is 110°, but that would mean angle BAD is 330°, which is impossible.

Alternatively, maybe the 110° and 100° are angles at A and D?

If angle BAD is 110°, and angle CDA is 100°, then as before, x + y = 50°, leading to angle AFD = 130°.

Alternatively, if angle BAD is 100°, angle CDA is 110°, then x + y = (100 + 110)/3 = 70°, so angle AFD = 180 - 70 = 110°.

But without knowing which angle is which, it's ambiguous.

However, looking back at the Asymptote code, there are labels for x and y in different positions. The label "x" is near (1, 1.5), (2, 1), (2, -1). Let me think about what those positions might represent.

Point (1, 1.5) is near the middle of AB. If angle BAD is trisected, then the first trisection angle from AB would be near that area. Similarly, (2, 1) is closer to B, and (2, -1) is below.

Similarly, labels for y are near (5.5, 0.5), (6.5, 0), (5, -1.5). These are near the middle of CD and around D.

If angle BAD is trisected, then the trisection angles would be near those x labels. Similarly for angle CDA.

If angle BAD is trisected into three angles, and one of them is labeled x, then angle BAD = 3x. Similarly, angle CDA = 3y.

If we can find another relation between x and y, we can solve for them.

Earlier, using the quadrilateral angle sum, if we assume that the labeled angles 110° and 100° are angles at B and C, then angle BAD + angle CDA = 3x + 3y = 360 - 110 - 100 - angle BCD. Wait, but we don't know angle BCD.

Wait, earlier calculation with coordinates gave angle BCD = 90°, but if the diagram labels it as 100°, then angle BCD = 100°, and angle ABC = 110°, then angle BAD + angle CDA = 360 - 110 - 100 - 90 = 60°? No, that can't be.

Wait, no. If angle ABC = 110°, angle BCD = 100°, then angle BAD + angle CDA = 360 - 110 - 100 = 150°, so 3x + 3y = 150°, so x + y = 50°, as before.

Then angle AFD = 180 - x - y = 130°.

This seems plausible.

Alternatively, if angle ABC = 100°, angle BCD = 110°, then angle BAD + angle CDA = 360 - 100 - 110 = 150°, same result.

Therefore, regardless of which angle is which, x + y = 50°, leading to angle AFD = 130°.

Therefore, the measure of angle AFD is 130 degrees.

But let me verify this with another approach.

Assume that angle BAD = 3x and angle CDA = 3y, with x + y = 50°.

If I can find another relation between x and y, I can confirm.

Alternatively, consider triangle AFD. If I can find the measures of angles at A and D, then I can find angle AFD.

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

Therefore, in triangle AFD, angles are x, y, and angle AFD.

So angle AFD = 180 - x - y = 180 - 50 = 130°.

Therefore, the answer is 130°.

But let me check with coordinates.

Earlier, with coordinates, angle BAD ≈ 79.4°, angle CDA ≈ 73.9°, so x ≈ 26.5°, y ≈ 24.6°, x + y ≈ 51.1°, so angle AFD ≈ 180 - 51.1 ≈ 128.9°, which is close to 130°, considering rounding errors.

Therefore, likely the answer is 130°.

But let me think again. If the problem labels 110° and 100°, which are likely angles at B and C, then angle BAD + angle CDA = 150°, leading to x + y = 50°, so angle AFD = 130°.

Therefore, I think the answer is 130 degrees.

**Final Answer**
\boxed{130}

To determine the measure of angle $ \angle AFD $ in quadrilateral $ABCD$, we are given that angles $ \angle BAD $ and $ \angle CDA $ are trisected. The diagram also includes labels for angles $ x $ and $ y $, and specific angle measures of $ 110^\circ $ and $ 100^\circ $, which are likely the measures of angles $ \angle ABC $ and $ \angle BCD $, respectively.

---

### Step 1: Use the Sum of Interior Angles in a Quadrilateral

In any quadrilateral, the sum of the interior angles is always:

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

From the diagram, we are told:

- $ \angle ABC = 110^\circ $
- $ \angle BCD = 100^\circ $

Let’s denote:

- $ \angle BAD = 3x $ (since it is trisected into three equal parts)
- $ \angle CDA = 3y $ (since it is also trisected into three equal parts)

Substituting into the angle sum equation:

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

$$
3x + 3y = 150^\circ
$$

$$
x + y = 50^\circ
$$

---

### Step 2: Analyze Triangle $ \triangle AFD $

We are asked to find the measure of $ \angle AFD $. In triangle $ \triangle AFD $, the angles at vertices $ A $ and $ D $ are:

- $ \angle FAD = x $ (since $ AF $ is a trisector of $ \angle BAD $)
- $ \angle FDA = y $ (since $ FD $ is a trisector of $ \angle CDA $)

Using the triangle angle sum property:

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

Substitute $ x + y = 50^\circ $:

$$
\angle AFD = 180^\circ - 50^\circ = \boxed{130^\circ}
$$

---

### Final Answer

$$
\boxed{130}
$$

[Assistant]
Explanation:
Okay, so I need to find the measure of angle AFD in this quadrilateral ABCD. The problem says that angles BAD and CDA are trisected, and there's an Asymptote figure provided. Let me try to visualize this first.

First, let me recall what trisecting an angle means. Trisecting an angle means dividing it into three equal parts. So, if angle BAD is trisected, there will be two lines from point A that divide angle BAD into three equal angles. Similarly, angle CDA is trisected, so there will be two lines from point D that divide angle CDA into three equal angles. The figure shows points E and F, which are probably the trisection points.

Looking at the Asymptote code, points E and F are defined. E is near (4.5 - 0.2, 1 - 0.2), which is approximately (4.3, 0.8), and F is at (5, 3). The labels for angles x and y are placed in different positions. There's also a label for 110 degrees near (2.5, 3.5) and 100 degrees near (6.5 - 0.2, 3.5).

Wait, maybe I should try to figure out the measures of angles BAD and CDA first? Because if they are trisected, then knowing those angles would help me find the trisection angles.

But how do I find angles BAD and CDA? The Asymptote code gives coordinates for points A, B, C, D. Let me try to use coordinates to calculate the angles.

Coordinates:
A = (0,0)
B = (2,4)
C = (7,4)
D = (7,-2)

So quadrilateral ABCD has coordinates A(0,0), B(2,4), C(7,4), D(7,-2). Let me sketch this mentally. From A(0,0) to B(2,4) to C(7,4) to D(7,-2) back to A.

First, let me compute angle BAD. That's the angle at point A between BA and DA. Similarly, angle CDA is the angle at point D between CD and DA.

To compute angle BAD, I can find vectors AB and AD and then compute the angle between them. Similarly for angle CDA.

First, vector AB is from A(0,0) to B(2,4): so vector AB is (2,4). Vector AD is from A(0,0) to D(7,-2): so vector AD is (7,-2).

The angle between AB and AD is angle BAD. Let me compute that using the dot product formula.

The formula for the angle θ between vectors u and v is:

cosθ = (u · v)/(|u| |v|)

So, compute AB · AD = (2)(7) + (4)(-2) = 14 - 8 = 6

|AB| = sqrt(2² + 4²) = sqrt(4 + 16) = sqrt(20) = 2*sqrt(5)

|AD| = sqrt(7² + (-2)²) = sqrt(49 + 4) = sqrt(53)

Therefore, cosθ = 6 / (2√5 * √53) = 6 / (2√265) = 3 / √265 ≈ 3 / 16.2788 ≈ 0.1844

Then θ ≈ arccos(0.1844) ≈ 80 degrees? Wait, let me check with calculator.

Wait, arccos(0.1844). Let me recall that cos(80°) ≈ 0.1736, cos(75°) ≈ 0.2588. So 0.1844 is between 75° and 80°, closer to 80°. Let me compute it more accurately.

Let me use a calculator. Let me compute arccos(3 / sqrt(265)).

First, sqrt(265) ≈ 16.2788, so 3 / 16.2788 ≈ 0.1844.

Now, arccos(0.1844). Let me convert this to degrees.

Using a calculator, arccos(0.1844) ≈ 80 degrees? Let me check with calculator steps.

Alternatively, maybe I can compute it using inverse cosine. But since I don't have a calculator here, maybe I can use approximate values.

Alternatively, maybe there's another way. Wait, but maybe I can find angle BAD using coordinates. Alternatively, maybe the problem gives some angles? Wait, the Asymptote code labels 110° and 100°. Let me check the labels.

The Asymptote code labels "110°" at (2.5, 3.5) and "100°" at (6.5 - 0.2, 3.5). Let me see where those points are.

Point (2.5, 3.5) is near the middle of AB and BC? Wait, AB is from (0,0) to (2,4). The midpoint of AB is (1,2). But (2.5, 3.5) is near point B(2,4) and C(7,4). Wait, maybe that's angle ABC? Or angle at B? Wait, the label is near (2.5, 3.5), which is above AB. Maybe that's angle ABC? Wait, but angle ABC is at point B. Let me check coordinates.

Alternatively, maybe those labels are for angles that are parts of the trisection? Wait, the problem says angles BAD and CDA are trisected. The Asymptote code labels "x" and "y" in different positions. Maybe those are the trisection angles?

Wait, the problem says "angle BAD and angle CDA are trisected as shown." So maybe the diagram shows trisection lines, and the labels x and y are the measures of those trisection angles? But the problem also labels 110° and 100°, which might be angles at B and C?

Alternatively, maybe the 110° and 100° are angles that are part of the trisection? Wait, the Asymptote code labels "110°" near (2.5, 3.5) which is near the middle of AB and BC. Maybe that's angle ABC? Let me check coordinates.

Point B is at (2,4). The angle at B is between AB and BC. Let me compute angle ABC.

Vector BA is from B to A: (-2, -4). Vector BC is from B to C: (5, 0).

So angle between BA and BC. Let me compute that.

Dot product: (-2)(5) + (-4)(0) = -10 + 0 = -10

|BA| = sqrt(4 + 16) = sqrt(20) = 2√5

|BC| = sqrt(25 + 0) = 5

So cosθ = (-10)/(2√5 * 5) = (-10)/(10√5) = -1/√5 ≈ -0.4472

Therefore θ ≈ 116.565 degrees. Wait, but the label says 110°. Hmm, maybe my calculation is wrong? Wait, maybe I need to check again.

Wait, angle ABC is at point B. The vectors are BA and BC. Wait, BA is from B to A, which is (-2, -4). BC is from B to C, which is (5, 0). So the angle between BA and BC is indeed calculated as above.

But according to calculation, it's arccos(-1/√5) ≈ 116.565°, but the diagram labels it as 110°. Maybe there's a discrepancy here? Or maybe my coordinates are wrong? Wait, the Asymptote code defines A(0,0), B(2,4), C(7,4), D(7,-2). So coordinates are correct.

Alternatively, maybe the diagram is not to scale? Because the problem says "as shown," but since I don't have the image, maybe the labels 110° and 100° are given for angles at B and C? Wait, the label "110°" is at (2.5, 3.5), which is near the middle of AB and BC. Maybe that's angle ABC? But according to calculation, it's about 116.5°, not 110°. Maybe the diagram is approximate?

Alternatively, maybe the 110° and 100° are angles that are parts of the trisection? Wait, the problem says angles BAD and CDA are trisected. 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 are those trisection angles?

Looking at the Asymptote code, there are labels for x in positions (1, 1.5), (2, 1), (2, -1). Maybe those are the trisection angles? Similarly, labels for y near (5.5, 0.5), (6.5, 0), (5, -1.5).

But without seeing the diagram, it's a bit challenging. However, maybe I can proceed with the coordinates.

Alternatively, maybe the problem is from a known source? Wait, the Asymptote code is given, so maybe I can reconstruct the diagram.

Let me try to think again. The quadrilateral ABCD with coordinates A(0,0), B(2,4), C(7,4), D(7,-2). So AB is from (0,0) to (2,4), BC is from (2,4) to (7,4), CD is from (7,4) to (7,-2), and DA is from (7,-2) back to (0,0).

So angle BAD is at point A between BA and DA. As calculated earlier, angle BAD ≈ 80 degrees? Wait, earlier calculation gave angle BAD ≈ arccos(3 / sqrt(265)) ≈ 80 degrees? Let me check with approximate values.

Alternatively, maybe I can compute the slopes of AB and AD to find the angle between them.

Slope of AB: from A(0,0) to B(2,4): rise over run is 4/2 = 2. So slope is 2.

Slope of AD: from A(0,0) to D(7,-2): rise over run is -2/7 ≈ -0.2857.

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

tanθ = |(m2 - m1)/(1 + m1*m2)|

So tanθ = |(-2/7 - 2)/(1 + (2)(-2/7))| = |(-2/7 - 14/7)/(1 - 4/7)| = |(-16/7)/(3/7)| = | -16/3 | = 16/3 ≈ 5.333...

Therefore θ = arctan(16/3) ≈ arctan(5.333). Let me recall that tan(79°) ≈ 5.145, tan(80°) ≈ 5.671. So 16/3 ≈ 5.333 is between 79° and 80°, closer to 80°. Let me compute it more accurately.

Let me use linear approximation. Let’s say tan(79°) ≈ 5.145, tan(80°) ≈ 5.671. The difference between tan(80°) and tan(79°) is approximately 0.526 per degree. Our value is 5.333, which is 5.333 - 5.145 = 0.188 above tan(79°). So fraction is 0.188 / 0.526 ≈ 0.357. So angle ≈ 79° + 0.357° ≈ 79.357°, approximately 79.4 degrees.

So angle BAD ≈ 79.4 degrees. If this is trisected, then each trisection angle would be approximately 79.4 / 3 ≈ 26.5 degrees.

Similarly, angle CDA is at point D between CD and DA. Let me compute that angle.

Point D is at (7,-2). CD is from D(7,-2) to C(7,4), which is straight up along the vertical line x=7. DA is from D(7,-2) to A(0,0).

So vector DC is (0,6) (from D to C), and vector DA is (-7, 2) (from D to A).

So angle between DC and DA is angle CDA. Let me compute that.

Vectors DC = (0,6) and DA = (-7, 2).

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

|DC| = sqrt(0² + 6²) = 6

|DA| = sqrt((-7)^2 + 2^2) = sqrt(49 + 4) = sqrt(53) ≈ 7.28

Therefore cosθ = 12 / (6 * sqrt(53)) = 2 / sqrt(53) ≈ 2 / 7.28 ≈ 0.275

Therefore θ ≈ arccos(0.275) ≈ 73.9 degrees.

So angle CDA ≈ 73.9 degrees. If trisected, each trisection angle would be ≈ 73.9 / 3 ≈ 24.6 degrees.

But the problem mentions that angles BAD and CDA are trisected, and there are labels for x and y. However, in the Asymptote code, there are labels for x and y in different positions. Maybe those x and y are the trisection angles?

But the problem also labels 110° and 100°, which might be angles at B and C? Wait, earlier calculation for angle ABC was approximately 116.5°, but the diagram labels it as 110°, so maybe there's inconsistency.

Alternatively, maybe the given angles of 110° and 100° are the measures of angles BAD and CDA? Wait, if angle BAD is 110°, then trisected into three angles of 110/3 ≈ 36.66°, and angle CDA is 100°, trisected into 100/3 ≈ 33.33°. But earlier calculations with coordinates gave angle BAD ≈ 79.4° and angle CDA ≈ 73.9°, which don't match 110° and 100°.

This suggests that maybe the diagram is not drawn to scale, or perhaps the labels 110° and 100° are not angles BAD and CDA but something else.

Alternatively, maybe the 110° and 100° are angles formed by the trisection lines?

Wait, the Asymptote code draws A--E--D and A--F--D. So E and F are points on AD? Wait, no. Wait, E is defined as (4.5 - 0.2, 1 - 0.2) which is approximately (4.3, 0.8). Let me check if that's on any line.

Looking at the coordinates, point E is near (4.3, 0.8). Let me see if that's on AB or AD or BC or CD?

Point A is (0,0), B is (2,4), C is (7,4), D is (7,-2).

If I consider the line from A(0,0) to E(4.3, 0.8), that would be a line from A to somewhere inside the quadrilateral. Similarly, line from A to F(5,3).

But the Asymptote code draws A--E--D and A--F--D. So E and F are points such that AE and AF are trisectors of angle BAD, and DE and DF are trisectors of angle CDA?

Wait, the problem says angles BAD and CDA are trisected. So angle BAD is trisected, meaning there are two lines from A dividing angle BAD into three equal parts. Similarly, angle CDA is trisected, meaning two lines from D dividing angle CDA into three equal parts.

In the diagram, points E and F are drawn. The Asymptote code draws A--E--D and A--F--D. So E and F are points on AD? Wait, no. If you draw A--E--D, then E is a point such that AE and ED are parts of the trisection? Wait, no. If angle BAD is trisected, then the trisectors would be lines from A, not from D. Similarly, angle CDA is trisected, so trisectors from D.

Wait, maybe E is on AB and F is on DC? Or somewhere else?

Alternatively, maybe E and F are points where the trisectors intersect some sides?

This is getting confusing without seeing the diagram. Maybe I need to think differently.

The problem asks for the measure of angle AFD. Points A, F, D. So angle at F between A and D.

Given that F is a trisection point of angle CDA? Or angle BAD?

Wait, the Asymptote code defines E and F as points. Let me check their coordinates.

E is defined as (4.5 - 0.2, 1 - 0.2) which is (4.3, 0.8). Let me see if that's on any line.

Looking at the coordinates, point B is (2,4), C is (7,4). The line BC is horizontal from (2,4) to (7,4). The point E is at (4.3, 0.8), which is below BC.

Alternatively, maybe E is on BD? Let me check.

Point B is (2,4), D is (7,-2). The line BD can be parametrized. Let me see if (4.3, 0.8) is on BD.

Parametric equations for BD: starting at B(2,4), direction vector (5, -6). So parametric equations: x = 2 + 5t, y = 4 - 6t.

Let me see if there's a t such that x = 4.3 and y = 0.8.

From x: 2 + 5t = 4.3 => 5t = 2.3 => t = 0.46

Then y = 4 - 6*(0.46) = 4 - 2.76 = 1.24. But the y-coordinate of E is 0.8, not 1.24. So not on BD.

Alternatively, maybe E is on AB? Let me check. AB is from (0,0) to (2,4). Let me parametrize AB: x = 2s, y = 4s, where s ∈ [0,1].

If E is (4.3, 0.8), then 2s = 4.3 => s = 2.15, which is outside the range [0,1]. So not on AB.

Similarly, check if E is on AD. AD is from (0,0) to (7,-2). Parametric equations: x = 7t, y = -2t.

If E is (4.3, 0.8), then 7t = 4.3 => t ≈ 0.614, then y = -2*(0.614) ≈ -1.228, but E's y-coordinate is 0.8. Not matching.

So E is not on AD. Maybe E is on BC? BC is from (2,4) to (7,4). So y-coordinate is 4. But E has y=0.8, so no.

Alternatively, maybe E is inside the quadrilateral?

Similarly, point F is at (5,3). Let me check if that's on any line.

Looking at coordinates, point F(5,3). Let me check if it's on AB, BC, CD, or DA.

AB: from (0,0) to (2,4). Parametric equations: x=2s, y=4s. If x=5, then s=2.5, which is outside.

BC: horizontal line y=4. F has y=3, so no.

CD: vertical line x=7 from (7,4) to (7,-2). F has x=5, so no.

DA: from (7,-2) to (0,0). Parametric equations: x=7-7t, y=-2 + 2t. If x=5, then 7 -7t =5 => t=2/7. Then y = -2 + 2*(2/7) = -2 + 4/7 ≈ -1.428, which is not 3.

So F is not on any side. Therefore, E and F are points inside the quadrilateral.

Given that, and the Asymptote code draws A--E--D and A--F--D. So AE and AF are lines from A to E and F, and ED and FD are lines from E and F to D.

But the problem says angles BAD and CDA are trisected. So angle BAD is trisected by AE and AF? Or maybe AE and another line?

Wait, angle BAD is at point A. If it's trisected, there should be two lines from A dividing angle BAD into three equal angles. Similarly, angle CDA is at point D, so two lines from D dividing angle CDA into three equal angles.

In the diagram, there are two lines from A: AE and AF. Maybe those are the trisectors of angle BAD? Similarly, maybe there are two lines from D, but in the diagram, only FD is drawn? Wait, the Asymptote code draws A--E--D and A--F--D. So from A, two lines AE and AF going to E and F, and then from E and F to D.

But if angle BAD is trisected, then AE and AF would be the trisectors. Similarly, if angle CDA is trisected, then maybe FD is one of the trisectors?

But the problem says both angles BAD and CDA are trisected. So maybe there are two trisectors for each angle.

But since the diagram shows only two lines from A (AE and AF) and one line from D (FD), maybe FD is a trisector of angle CDA?

Alternatively, maybe there are two trisectors from D, but only FD is shown?

This is getting too ambiguous without the diagram. Maybe I need to use the given labels.

The Asymptote code labels "x" in three places: (1, 1.5), (2, 1), (2, -1). Maybe those are the trisection angles of angle BAD? Similarly, labels "y" in three places: (5.5, 0.5), (6.5, 0), (5, -1.5). Maybe those are trisection angles of angle CDA?

If that's the case, then angle BAD is divided into three angles of measure x, and angle CDA is divided into three angles of measure y.

Additionally, there are labels "110°" and "100°". Maybe those are angles at B and C?

Alternatively, maybe the 110° and 100° are the measures of angles BAD and CDA?

Wait, if angle BAD is 110°, then each trisection angle is 110/3 ≈ 36.66°, and angle CDA is 100°, each trisection angle is 100/3 ≈ 33.33°.

But earlier calculations with coordinates gave angle BAD ≈ 79.4° and angle CDA ≈ 73.9°, which don't match.

Alternatively, maybe the 110° and 100° are angles formed by the trisectors?

Alternatively, maybe the problem is from a specific source where the diagram is standard, and the angles BAD and CDA are 110° and 100°, which are trisected.

Wait, the problem says "angle BAD and angle CDA are trisected as shown." So maybe the diagram shows that angle BAD is trisected into three angles, and angle CDA is trisected into three angles, and the labels 110° and 100° are for those angles?

But if angle BAD is trisected, then the total angle would be 3x, and angle CDA would be 3y.

But the problem doesn't give specific values for x or y, but labels them as x and y. However, there are also labels for 110° and 100°, which might be angles at B and C?

Alternatively, maybe the 110° and 100° are the measures of angles BAD and CDA?

If angle BAD is 110°, then each trisection angle is 110/3 ≈ 36.66°, and angle CDA is 100°, each trisection angle is 100/3 ≈ 33.33°.

But then, how does that help me find angle AFD?

Alternatively, maybe the diagram is such that angle BAD is 110°, angle CDA is 100°, and they are trisected, so we can find the measure of angle AFD using those trisection angles.

Let me assume that angle BAD = 110°, trisected into three angles of 110/3 ≈ 36.66°, and angle CDA = 100°, trisected into three angles of 100/3 ≈ 33.33°.

Then, points E and F are the trisection points.

But how does that help me find angle AFD?

Let me think about triangle AFD. To find angle AFD, I need to know some information about triangle AFD. Maybe I can find the measures of angles at A and D in triangle AFD, then use the fact that the sum of angles in a triangle is 180°.

But angle at A in triangle AFD is part of angle BAD. Similarly, angle at D in triangle AFD is part of angle CDA.

Assuming that angle BAD is trisected, then if AF is one of the trisectors, then angle FAD would be one-third of angle BAD. Similarly, if FD is one of the trisectors of angle CDA, then angle FDA would be one-third of angle CDA.

But I need to figure out which trisectors AF and FD are.

Looking back at the Asymptote code, the label "x" is near (1, 1.5), (2, 1), (2, -1). Maybe those are the trisection angles of angle BAD. Similarly, labels "y" near (5.5, 0.5), (6.5, 0), (5, -1.5) are trisection angles of angle CDA.

If angle BAD is trisected into three angles of measure x, then angle BAD = 3x. Similarly, angle CDA = 3y.

Additionally, there are labels for 110° and 100°, which might be angles at B and C.

But without knowing the exact configuration, it's challenging. However, maybe I can use the fact that in quadrilateral ABCD, the sum of all interior angles is 360°.

Wait, quadrilateral ABCD has angles at A, B, C, D.

If I can find angles at A, B, C, D, then maybe I can find relations.

Earlier, I calculated angle BAD ≈ 79.4°, angle ABC ≈ 116.5°, angle BCD, and angle CDA ≈ 73.9°. Let me compute angle BCD.

Point C is at (7,4). Angle BCD is at point C between BC and CD.

Vector CB is from C to B: (-5, 0). Vector CD is from C to D: (0, -6).

So angle between CB and CD.

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

Wait, that means the angle is 90 degrees? Because the dot product is zero.

So angle BCD is 90 degrees.

Therefore, sum of angles in quadrilateral ABCD: angle A + angle B + angle C + angle D = 79.4 + 116.5 + 90 + 73.9 ≈ 359.8°, which is roughly 360°, considering rounding errors.

So angle BAD ≈ 79.4°, angle ABC ≈ 116.5°, angle BCD = 90°, angle CDA ≈ 73.9°.

But the problem labels 110° and 100°, which might be approximate values for angles at B and C?

If angle ABC is labeled as 110°, and angle BCD is labeled as 100°, but according to calculation, angle ABC is ≈ 116.5°, angle BCD is 90°, which doesn't match.

Alternatively, maybe the labels 110° and 100° are for angles BAD and CDA?

If angle BAD is 110°, angle CDA is 100°, then angle ABC and angle BCD can be calculated.

Let me try that.

Assume angle BAD = 110°, angle CDA = 100°, angle BCD = 90°, then angle ABC = 360 - 110 - 100 - 90 = 60°.

But according to coordinates, angle ABC was ≈ 116.5°, which conflicts.

Alternatively, maybe the diagram is different.

Alternatively, maybe the given angles of 110° and 100° are the measures of angles formed by the trisectors?

This is getting too confusing. Maybe I need to approach this problem using the trisection angles and the given labels.

Let me consider that angle BAD is trisected into three angles of measure x, and angle CDA is trisected into three angles of measure y.

From the Asymptote code, there are labels for x and y. Also, there are labels for 110° and 100°. Maybe those are angles that are formed by the trisectors?

Alternatively, maybe the 110° and 100° are angles at B and C, and using that, we can find x and y.

Let me try that.

Assume angle ABC = 110°, angle BCD = 100°, and angle BAD = 3x, angle CDA = 3y.

Then, in quadrilateral ABCD, sum of angles is 360°, so:

angle BAD + angle ABC + angle BCD + angle CDA = 3x + 110 + 100 + 3y = 360

So 3x + 3y = 360 - 110 - 100 = 150

Therefore, x + y = 50°

So x + y = 50°.

Now, need more equations to find x and y.

Alternatively, maybe using triangle AFD.

But how?

Alternatively, consider triangle AFD. To find angle AFD, we need to know some relations.

Assuming that AF is a trisector of angle BAD, so angle FAD = x, and FD is a trisector of angle CDA, so angle FDA = y.

Then, in triangle AFD, angles at A and D are x and y, so angle AFD = 180° - x - y = 180° - (x + y) = 180° - 50° = 130°.

Wait, that seems straightforward! If angle FAD = x and angle FDA = y, then angle AFD = 180 - x - y = 130°.

But is this valid?

Wait, but why would angle FAD = x and angle FDA = y?

If AF is a trisector of angle BAD, then angle FAD would be one-third of angle BAD, which is x. Similarly, if FD is a trisector of angle CDA, then angle FDA would be one-third of angle CDA, which is y.

But is FD a trisector of angle CDA?

In the diagram, FD is drawn from F to D. If angle CDA is trisected, then there should be two lines from D dividing angle CDA into three equal parts. If FD is one of those trisectors, then angle FDA would be y.

But how do we know that FD is a trisector?

Alternatively, maybe AF is a trisector of angle BAD, and FD is a trisector of angle CDA.

But without seeing the diagram, it's hard to be certain. However, given that the problem labels angles x and y, and given that angle BAD = 3x and angle CDA = 3y, and from the quadrilateral angle sum we found x + y = 50°, then angle AFD = 180 - x - y = 130°.

But the problem might have different configuration.

Wait, but if angle AFD is 130°, is that the answer?

But let me check with the coordinates.

Earlier, I calculated angle BAD ≈ 79.4°, angle CDA ≈ 73.9°, so x ≈ 26.5°, y ≈ 24.6°, so x + y ≈ 51.1°, so angle AFD ≈ 180 - 51.1 ≈ 128.9°, which is roughly 130°. So maybe the answer is 130°?

But the problem might have exact values. If angle BAD and CDA are trisected, and with the given labels of 110° and 100°, which might be angles at B and C, leading to x + y = 50°, then angle AFD = 130°.

Alternatively, maybe the labels 110° and 100° are for angles BAD and CDA?

If angle BAD = 110°, angle CDA = 100°, then x = 110/3 ≈ 36.66°, y = 100/3 ≈ 33.33°, so x + y ≈ 70°, angle AFD = 180 - 70 = 110°. But this contradicts previous calculation.

But how do I know which is correct?

Wait, the Asymptote code labels "110°" near (2.5, 3.5) and "100°" near (6.5 - 0.2, 3.5). Let me think about where those points are.

Point (2.5, 3.5) is near the middle of AB and BC? If AB is from (0,0) to (2,4), and BC is from (2,4) to (7,4). The point (2.5, 3.5) is between AB and BC. Maybe that's angle ABC? But earlier calculation said angle ABC ≈ 116.5°, but labeled as 110°.

Similarly, the label "100°" is near (6.5 - 0.2, 3.5) which is (6.3, 3.5). That's near point C(7,4). Maybe that's angle BCD? Earlier calculation said angle BCD is 90°, but labeled as 100°.

This inconsistency suggests that the diagram is not to scale, or the labels are for different angles.

Alternatively, maybe the 110° and 100° are the measures of angles formed by the trisectors?

For example, if angle BAD is trisected into three angles, and one of them is 110°, but that would mean angle BAD is 330°, which is impossible.

Alternatively, maybe the 110° and 100° are angles at A and D?

If angle BAD is 110°, and angle CDA is 100°, then as before, x + y = 50°, leading to angle AFD = 130°.

Alternatively, if angle BAD is 100°, angle CDA is 110°, then x + y = (100 + 110)/3 = 70°, so angle AFD = 180 - 70 = 110°.

But without knowing which angle is which, it's ambiguous.

However, looking back at the Asymptote code, there are labels for x and y in different positions. The label "x" is near (1, 1.5), (2, 1), (2, -1). Let me think about what those positions might represent.

Point (1, 1.5) is near the middle of AB. If angle BAD is trisected, then the first trisection angle from AB would be near that area. Similarly, (2, 1) is closer to B, and (2, -1) is below.

Similarly, labels for y are near (5.5, 0.5), (6.5, 0), (5, -1.5). These are near the middle of CD and around D.

If angle BAD is trisected, then the trisection angles would be near those x labels. Similarly for angle CDA.

If angle BAD is trisected into three angles, and one of them is labeled x, then angle BAD = 3x. Similarly, angle CDA = 3y.

If we can find another relation between x and y, we can solve for them.

Earlier, using the quadrilateral angle sum, if we assume that the labeled angles 110° and 100° are angles at B and C, then angle BAD + angle CDA = 3x + 3y = 360 - 110 - 100 - angle BCD. Wait, but we don't know angle BCD.

Wait, earlier calculation with coordinates gave angle BCD = 90°, but if the diagram labels it as 100°, then angle BCD = 100°, and angle ABC = 110°, then angle BAD + angle CDA = 360 - 110 - 100 - 90 = 60°? No, that can't be.

Wait, no. If angle ABC = 110°, angle BCD = 100°, then angle BAD + angle CDA = 360 - 110 - 100 = 150°, so 3x + 3y = 150°, so x + y = 50°, as before.

Then angle AFD = 180 - x - y = 130°.

This seems plausible.

Alternatively, if angle ABC = 100°, angle BCD = 110°, then angle BAD + angle CDA = 360 - 100 - 110 = 150°, same result.

Therefore, regardless of which angle is which, x + y = 50°, leading to angle AFD = 130°.

Therefore, the measure of angle AFD is 130 degrees.

But let me verify this with another approach.

Assume that angle BAD = 3x and angle CDA = 3y, with x + y = 50°.

If I can find another relation between x and y, I can confirm.

Alternatively, consider triangle AFD. If I can find the measures of angles at A and D, then I can find angle AFD.

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

Therefore, in triangle AFD, angles are x, y, and angle AFD.

So angle AFD = 180 - x - y = 180 - 50 = 130°.

Therefore, the answer is 130°.

But let me check with coordinates.

Earlier, with coordinates, angle BAD ≈ 79.4°, angle CDA ≈ 73.9°, so x ≈ 26.5°, y ≈ 24.6°, x + y ≈ 51.1°, so angle AFD ≈ 180 - 51.1 ≈ 128.9°, which is close to 130°, considering rounding errors.

Therefore, likely the answer is 130°.

But let me think again. If the problem labels 110° and 100°, which are likely angles at B and C, then angle BAD + angle CDA = 150°, leading to x + y = 50°, so angle AFD = 130°.

Therefore, I think the answer is 130 degrees.

**Final Answer**
\boxed{130}

To determine the measure of angle $ \angle AFD $ in quadrilateral $ABCD$, we are given that angles $ \angle BAD $ and $ \angle CDA $ are trisected. The diagram also includes labels for angles $ x $ and $ y $, and specific angle measures of $ 110^\circ $ and $ 100^\circ $, which are likely the measures of angles $ \angle ABC $ and $ \angle BCD $, respectively.

---

### Step 1: Use the Sum of Interior Angles in a Quadrilateral

In any quadrilateral, the sum of the interior angles is always:

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

From the diagram, we are told:

- $ \angle ABC = 110^\circ $
- $ \angle BCD = 100^\circ $

Let’s denote:

- $ \angle BAD = 3x $ (since it is trisected into three equal parts)
- $ \angle CDA = 3y $ (since it is also trisected into three equal parts)

Substituting into the angle sum equation:

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

$$
3x + 3y = 150^\circ
$$

$$
x + y = 50^\circ
$$

---

### Step 2: Analyze Triangle $ \triangle AFD $

We are asked to find the measure of $ \angle AFD $. In triangle $ \triangle AFD $, the angles at vertices $ A $ and $ D $ are:

- $ \angle FAD = x $ (since $ AF $ is a trisector of $ \angle BAD $)
- $ \angle FDA = y $ (since $ FD $ is a trisector of $ \angle CDA $)

Using the triangle angle sum property:

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

Substitute $ x + y = 50^\circ $:

$$
\angle AFD = 180^\circ - 50^\circ = \boxed{130^\circ}
$$

---

### Final Answer

$$
\boxed{130}
$$

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.