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.

Alternatively, maybe I can compute it using exact values? Wait, but maybe there's another way. Wait, but maybe the problem is designed so that angles BAD and CDA are trisected, and with the given labels of 110° and 100°, maybe those are angles at B and C? Wait, 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 those labels are angles at B and C?

Wait, the label 110° is near (2.5, 3.5). If I imagine the figure, point B is at (2,4), and point C is at (7,4). So (2.5, 3.5) is near the top of the quadrilateral. Maybe that's angle ABC? Similarly, 100° is near (6.5 - 0.2, 3.5), which is near point C(7,4). Maybe angle BCD?

But the problem says that angles BAD and CDA are trisected. So maybe angles at A and D are being trisected.

But perhaps I need to find angles BAD and CDA first. Let me try again.

Alternatively, maybe the given angles of 110° and 100° are the trisected angles? Wait, the Asymptote code labels "x" and "y" in different positions. Let me check the Asymptote code again:

label("$x$", (1, 1.5), S); label("$x$", (2, 1), S+W); label("$x$", (2, -1), N+N+N+W);
label("$y$", (5.5+.3, .5-.3), S); label("$y$", (6.5+.3, 0)); label("$y$", (5+.5, -1.5+.3));
label("$110^{\circ}$",(2.5,3.5)); label("$100^{\circ}$",(6.5-.2,3.5));

So, the labels for x are near the middle of AB? Maybe those are angles created by the trisection? Similarly for y.

But maybe the angles labeled 110° and 100° are angles at B and C? Let me think.

Alternatively, maybe angle ABC is 110°, angle BCD is 100°, and then we can use that to find angles BAD and CDA?

But since ABCD is a quadrilateral, the sum of all interior angles is (4-2)*180 = 360°. If I can find angles at B and C, then maybe I can find angles at A and D.

But let me check. If angle ABC is 110°, angle BCD is 100°, then angles at A and D would be?

But without knowing angles at A and D, how can I find them? Wait, but maybe we can compute them using coordinates.

Alternatively, maybe the angles labeled 110° and 100° are the trisected angles? Wait, the problem says that angles BAD and CDA are trisected. So maybe those angles are being split into three equal parts.

But the Asymptote code shows that there are trisectors from A and D. For angle BAD, there are two trisectors, so three angles. Similarly for angle CDA.

But the labels for x and y are near the trisectors. For example, the first x is at (1, 1.5), which is near the middle of AB? Maybe that's one of the trisection angles?

Alternatively, maybe the angles labeled x and y are the trisection angles. For example, if angle BAD is trisected into three angles of measure x each, then angle BAD = 3x. Similarly, angle CDA is trisected into three angles of measure y each, so angle CDA = 3y.

But then, if that's the case, then maybe we can find x and y?

But how? The problem also labels 110° and 100°, which might be other angles in the figure.

Alternatively, maybe the angles labeled 110° and 100° are angles at B and C? Let me check.

If I can compute angles at B and C using coordinates, then maybe I can find angles at A and D.

Let me compute angle ABC.

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

Vector BA is from B to A: A - B = (0 - 2, 0 - 4) = (-2, -4)

Vector BC is from B to C: C - B = (7 - 2, 4 - 4) = (5, 0)

So angle ABC is the angle between vectors BA (-2, -4) and BC (5, 0).

Compute the angle between these vectors.

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

|BA| = sqrt((-2)^2 + (-4)^2) = sqrt(4 + 16) = sqrt(20) = 2√5

|BC| = sqrt(5^2 + 0^2) = 5

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

Therefore θ ≈ arccos(-0.4472) ≈ 116.565°. Wait, but the Asymptote code labels 110° near that area. Hmm, discrepancy here. Maybe my calculation is wrong?

Wait, let me check again.

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

Yes. The angle between BA and BC is angle ABC.

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

|BA| is sqrt(4 + 16) = sqrt(20). |BC| is 5.

So cosθ = -10 / (sqrt(20)*5) = -10 / (5*sqrt(20)) = -2 / sqrt(20) = -2 / (2*sqrt(5)) = -1/sqrt(5) ≈ -0.4472.

Yes, arccos(-1/√5) is approximately 116.565°, which is roughly 116.57°, but the Asymptote code labels 110° there. Maybe there's a mistake? Or maybe my assumption about which angle is labeled is wrong?

Alternatively, maybe the angle labeled 110° is not angle ABC? Let me check the coordinates of the label. The label is at (2.5, 3.5). Point B is at (2,4), so (2.5, 3.5) is slightly to the right and down from B. Maybe that's angle ABD or something else?

Alternatively, maybe the angle labeled 110° is angle BAD? Wait, angle BAD is at point A. But the label is at (2.5, 3.5), which is near the top of the figure.

Alternatively, maybe the 110° is angle ADC? Wait, angle ADC is at D.

Alternatively, maybe the 110° is angle AED or something else?

This is getting confusing. Maybe I need to approach this differently.

Alternatively, since the problem mentions that angles BAD and CDA are trisected, and there are points E and F which are the trisection points. The Asymptote code draws A--E--D and A--F--D. So E and F are points on AD? Wait, no. Wait, A is connected to E and E is connected to D. So E is somewhere inside the quadrilateral? Similarly, F is connected from A to F to D.

Wait, looking at the coordinates of E and F.

E is defined as (4.5 - 0.2, 1 - 0.2) which is (4.3, 0.8). F is (5, 3).

So point E is somewhere near the middle of the figure? Let me check coordinates.

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

So E is at (4.3, 0.8). That seems to be inside the quadrilateral. Similarly, F is at (5,3), which is also inside.

But how are E and F related to the trisection?

The Asymptote code draws A--E--D and A--F--D. So from A, there are two lines: one to E and one to F, both going to D. So E and F are points on AD? Wait, but if you connect A to E to D, then E must be on AD? But if E is on AD, then the trisection would be along AD? But angle BAD is at A, so trisecting angle BAD would involve lines from A, not from D.

Wait, maybe E is on BD or something else?

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

Wait, the problem says "angle BAD and angle CDA are trisected as shown." The Asymptote code draws A--E--D and A--F--D. So from A, two lines are drawn to E and F, which then connect to D. So maybe E and F are points on BD or CD?

Alternatively, maybe E is on BC and F is on CD?

But without seeing the figure, it's a bit challenging. However, since the Asymptote code is given, maybe I can reconstruct the figure.

The Asymptote code draws the quadrilateral A(0,0)-B(2,4)-C(7,4)-D(7,-2)-cycle. Then draws A--E--D and A--F--D. Points E and F are defined as E = (4.5 - 0.2, 1 - 0.2) which is (4.3, 0.8) and F = (5, 3).

So plotting these points, E is somewhere near the middle of the figure, and F is higher up.

But how do these points relate to trisection?

If angle BAD is trisected, then from point A, there should be two lines dividing angle BAD into three equal angles. Similarly, from point D, two lines dividing angle CDA into three equal angles.

But in the Asymptote code, they draw A--E--D and A--F--D. So from A, two lines to E and F, which then go to D. So E and F are points on AD? Wait, but if E is on AD, then connecting A to E to D would just be a straight line. But E is at (4.3, 0.8), and AD goes from (0,0) to (7,-2). Let me check if E is on AD.

Parametrize AD: from A(0,0) to D(7,-2). The parametric equations are x = 7t, y = -2t, where t ∈ [0,1].

For E(4.3, 0.8), let's see if there exists t such that 7t = 4.3 and -2t = 0.8.

From y-coordinate: -2t = 0.8 ⇒ t = -0.4. But t should be between 0 and 1. So E is not on AD. Similarly for F(5,3). Let me check if F is on AD.

For x = 5 = 7t ⇒ t = 5/7 ≈ 0.714. Then y should be -2*(5/7) ≈ -1.428. But F has y-coordinate 3, which is not equal to -1.428. So F is not on AD.

Therefore, E and F are not on AD. Therefore, the lines AE and AF are trisectors of angle BAD, and similarly, maybe DE and DF are trisectors of angle CDA? Wait, but the Asymptote code draws A--E--D and A--F--D. So E and F are connected to D.

Wait, maybe E and F are points where the trisectors meet other sides? For example, trisectors of angle BAD might meet BC or CD?

Alternatively, since angle BAD is at A, trisectors would be lines from A, so AE and AF are the trisectors. Similarly, trisectors of angle CDA would be lines from D, say, DG and DH, which might meet AB or BC?

But the Asymptote code only shows A--E--D and A--F--D. Maybe E and F are the trisection points of angle BAD, and then lines from D trisecting angle CDA? But the figure only shows lines from A to E and F to D.

Alternatively, maybe E and F are the trisection points of angles BAD and CDA respectively?

This is getting too confusing without seeing the figure. Maybe I need to use the given labels.

The Asymptote code labels "x" in three places: (1, 1.5), (2, 1), (2, -1). These might be angles created by the trisection. Similarly, "y" is labeled in three places near the other side.

Also, there are labels of 110° and 100°. Maybe those are angles that can help us find the measures.

Alternatively, since the problem is about trisection, maybe the angles BAD and CDA can be determined using the given angles?

Wait, if angle BAD is trisected, and angle CDA is trisected, and there are some angles labeled as x and y, maybe we can set up equations.

Alternatively, since the problem is asking for angle AFD, which is the angle at F between A and D. So points A, F, D.

Given that F is connected from A to F to D, so F is somewhere inside the quadrilateral.

Alternatively, maybe using the coordinates of F(5,3) and A(0,0) and D(7,-2), we can compute angle AFD?

Wait, if I can find coordinates of F, then I can compute vectors FA and FD, then compute the angle between them.

But wait, the Asymptote code defines F as (5,3). Is that exact? Let me check.

In the Asymptote code, F is defined as (5,3). So if that's the case, then coordinates are given. So maybe I can compute angle AFD using coordinates?

But the problem says "as shown," so maybe the diagram is to scale? But since it's an Asymptote code, maybe the coordinates are exact?

Wait, but if I can use coordinates, then maybe I can compute angle AFD directly.

Let me try that.

Point A is (0,0), F is (5,3), D is (7,-2).

So vectors FA and FD.

Vector FA is from F to A: A - F = (0 - 5, 0 - 3) = (-5, -3)

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

Then angle AFD is the angle between vectors FA and FD.

Compute the angle between (-5, -3) and (2, -5).

Using the dot product formula:

cosθ = ( (-5)(2) + (-3)(-5) ) / ( |FA| |FD| )

Compute numerator: (-10) + 15 = 5

|FA| = sqrt((-5)^2 + (-3)^2) = sqrt(25 + 9) = sqrt(34)

|FD| = sqrt(2^2 + (-5)^2) = sqrt(4 + 25) = sqrt(29)

Therefore, cosθ = 5 / (sqrt(34)*sqrt(29)) ≈ 5 / (5.830 * 5.385) ≈ 5 / 31.35 ≈ 0.1595

Then θ ≈ arccos(0.1595) ≈ 80.9 degrees.

But the problem is likely expecting an exact value, not an approximate decimal. So maybe my approach is wrong? Because if I use coordinates, I get approximately 81 degrees, but maybe the exact value is something else.

But why would the problem give labels of 110° and 100°? Maybe those are angles that can help us find the exact measure?

Alternatively, maybe the coordinates are not exact? Because in the Asymptote code, E is defined as (4.5 - 0.2, 1 - 0.2), which is (4.3, 0.8), and F is (5, 3). Maybe these coordinates are approximate?

Alternatively, maybe the diagram is not to scale, and I need to solve it using angle trisection and given angles.

Let me try that approach.

Let me assume that angles BAD and CDA are trisected. Let me denote angle BAD = 3x and angle CDA = 3y.

The problem labels angles with x and y, so maybe those are the trisection angles?

In the Asymptote code, there are three labels of x near the middle of the figure, and three labels of y near the other side.

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

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

Since ABCD is a quadrilateral, the sum of its interior angles is 360°. So:

angle BAD + angle ABC + angle BCD + angle CDA = 360°

If angle ABC is 110°, angle BCD is 100°, then:

3x + 110° + 100° + 3y = 360°

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

Therefore, x + y = 50°

So if I can find another equation relating x and y, I can solve for them.

But how?

Looking back at the Asymptote code, there are labels for x and y. Maybe there are other angles that can be related.

Alternatively, maybe triangle AFD can be analyzed.

But without knowing more angles, it's difficult.

Alternatively, maybe considering triangle AFD, and using the trisection angles.

Wait, points E and F are trisection points. From the Asymptote code, A is connected to E and F, which are connected to D. So AE and AF are trisectors of angle BAD, and DE and DF are trisectors of angle CDA?

Wait, but angle CDA is at D, so trisectors would be from D. However, in the Asymptote code, they draw A--E--D and A--F--D. So maybe E and F are points where the trisectors of angle BAD meet some sides, and similarly for angle CDA?

Alternatively, maybe E and F are the trisection points of angles BAD and CDA, respectively.

But since angle BAD is trisected, there are two trisectors, so maybe AE and AF are those trisectors. Similarly, for angle CDA, there are two trisectors, but the Asymptote code only shows lines from A to E and F to D.

This is getting too vague. Maybe I need to use the given angles of 110° and 100°.

Assuming angle ABC = 110°, angle BCD = 100°, then angle BAD + angle CDA = 150°, as above.

If I can find another relationship between x and y, then I can find x and y.

Alternatively, maybe considering triangle AFD.

But to find angle AFD, maybe I need to find some relations.

Alternatively, since AE and AF trisect angle BAD, then angle BAE = angle EAF = angle FAD = x. Similarly, if DF and DE trisect angle CDA, then angle ADC is trisected into three angles of y.

But without knowing where E and F are located, it's hard to say.

Wait, the Asymptote code labels "x" near the middle of AB and other places. Maybe those x's are the trisection angles.

Alternatively, maybe the angles labeled x are the angles between the trisectors.

Alternatively, since the problem is from an Olympiad or competition, maybe there's a trick.

Wait, if angle BAD is trisected into three angles of x, and angle CDA is trisected into three angles of y, and we have x + y = 50°, as above.

If I can find another equation involving x and y, then I can solve for them.

Alternatively, maybe considering triangle AFD.

But how?

Alternatively, maybe considering the lines AE and AF trisecting angle BAD, and lines from D trisecting angle CDA.

But without knowing where those lines intersect, it's hard.

Alternatively, maybe using the fact that in triangle AFD, the sum of angles is 180°, but I need more information.

Alternatively, maybe using the Law of Sines or Law of Cosines in some triangles.

Alternatively, since coordinates are given, maybe I can calculate the exact value of angle AFD.

Earlier, I calculated angle AFD ≈ 80.9°, which is close to 81°, but maybe it's exactly 80° or 81°?

Wait, let me recalculate the exact value.

cosθ = 5 / (sqrt(34)*sqrt(29))

Let me compute sqrt(34)*sqrt(29) = sqrt(34*29) = sqrt(986).

So cosθ = 5 / sqrt(986). Let me compute 5^2 / 986 = 25 / 986 ≈ 0.02536.

Wait, no, cosθ = 5 / sqrt(986). Let me compute sqrt(986) ≈ 31.4, so 5 / 31.4 ≈ 0.1592.

Then arccos(0.1592) ≈ 80.9°, as before.

But maybe there's an exact value?

Alternatively, maybe the diagram is constructed with specific angles.

Wait, if angle BAD is trisected, and angle CDA is trisected, and given that angle ABC is 110°, angle BCD is 100°, then angle BAD + angle CDA = 150°, so 3x + 3y = 150°, so x + y = 50°.

If I can find another relation between x and y, maybe from triangle AFD or something else.

Alternatively, maybe considering triangle AFD.

But to do that, I need to know some sides or other angles.

Alternatively, maybe considering the lines AE and AF trisecting angle BAD.

If angle BAD = 3x, then each trisection angle is x. Similarly, angle CDA = 3y.

If I can find relations between these angles and other angles in the figure.

Alternatively, maybe considering triangle AED or AFD.

Alternatively, since the Asymptote code labels angles with x and y, maybe those are the trisection angles.

For example, if angle BAE = x, angle EAF = x, angle FAD = x, then angle BAD = 3x. Similarly, angle CDA = 3y.

If I can find relations between x and y using the other angles.

But how?

Alternatively, maybe using the fact that the sum of angles around point A and D.

Wait, at point A, the angles are angle BAD = 3x, and other angles?

Wait, in quadrilateral ABCD, at point A, we have angle BAD, and at point D, angle ADC.

But without more information, it's hard.

Alternatively, maybe using the given angles of 110° and 100°.

If angle ABC = 110°, angle BCD = 100°, then angle BAD + angle CDA = 150°, as before.

If I can find another equation involving x and y.

Wait, maybe considering triangle ABD or something else.

Alternatively, maybe considering the lines AE and AF.

Since AE and AF trisect angle BAD, then maybe there are some relations with triangle AED or AFD.

Alternatively, since the Asymptote code labels "x" near the middle of AB and other places, maybe those x's are angles that can be related to other angles.

Alternatively, maybe using the Law of Sines in triangle AFD.

But without knowing any sides, it's difficult.

Alternatively, maybe the answer is 80°, which is close to my approximate calculation. But how to confirm?

Alternatively, maybe there's a property when angles are trisected.

Wait, if angle BAD is trisected into three angles of x, and angle CDA is trisected into three angles of y, and x + y = 50°, then maybe angle AFD is related to x and y.

Alternatively, maybe angle AFD = 180° - (x + y) = 180° - 50° = 130°? But that seems too high.

Alternatively, maybe angle AFD = x + y = 50°? But that seems too low.

Alternatively, maybe angle AFD = 180° - 2x - 2y? Not sure.

Alternatively, maybe using the fact that in triangle AFD, the angles at A and D are related to x and y.

At point A, angle FAD = x (since it's one of the trisection angles). At point D, angle FDA is part of the trisection of angle CDA. If angle CDA is trisected into three angles of y, then angle ADC = 3y. If FD is one of the trisectors, then angle FDA = y?

Wait, if FD is a trisector of angle CDA, then angle ADC is split into three angles of y. So if FD is one of the trisectors, then angle FDC = y, and angle FDA = y? Wait, no.

Angle CDA is at point D, between CD and DA. If we trisect it, then from point D, two lines are drawn dividing angle CDA into three equal angles. Let's say those lines are DE and DF, making angles of y each. So angle CDE = y, angle EDF = y, angle FDA = y.

Therefore, angle FDA = y.

Similarly, at point A, angle FAD = x.

Therefore, in triangle AFD, we have angles at A: x, at D: y, and angle at F: angle AFD = 180° - x - y.

But earlier, we had x + y = 50°, so angle AFD = 180° - 50° = 130°.

Wait, that seems promising!

But why? Let me verify.

If angle FAD = x (since it's one of the trisection angles of angle BAD = 3x), and angle FDA = y (since it's one of the trisection angles of angle CDA = 3y), then in triangle AFD, the third angle at F is 180° - x - y.

But earlier, we found that x + y = 50°, so angle AFD = 180° - 50° = 130°.

But earlier, using coordinates, I found approximately 80.9°, which conflicts with this.

So there must be a mistake in my reasoning.

Wait, why did I assume that angle FDA = y?

If angle CDA is trisected, then from point D, two lines are drawn dividing angle CDA into three equal parts. If FD is one of those trisectors, then angle between DC and DF is y, and between DF and DA is y. So angle FDA would be y.

But in that case, angle FDA = y.

Similarly, angle FAD = x.

Therefore, in triangle AFD, angles at A and D are x and y, so angle AFD = 180 - x - y.

But earlier, we had x + y = 50°, so angle AFD = 130°.

But according to coordinates, it's approximately 80.9°, which is different.

This inconsistency suggests that my assumption about angle FDA = y is wrong.

Alternatively, maybe angle FDA is not y, but something else.

Wait, let me think again.

If angle CDA is trisected, then from point D, there are two trisectors. Let's say they divide angle CDA into three angles of measure y each. So angle between DC and first trisector is y, between first and second trisector is y, and between second trisector and DA is y.

If FD is one of those trisectors, then depending on which trisector it is, angle FDA could be y or 2y.

But in the Asymptote code, F is connected from A to F to D. So FD is a line from F to D. If F is a trisection point of angle BAD, then FD is not necessarily a trisector of angle CDA.

Wait, this is the confusion. Earlier, I assumed that FD is a trisector of angle CDA, but maybe it's not.

The Asymptote code draws A--E--D and A--F--D. So E and F are points connected from A to E and A to F, then to D. So AE and AF are trisectors of angle BAD, and ED and FD are...? Maybe they are trisectors of angle ADC?

But angle ADC is at D. If ED and FD are trisectors of angle ADC, then from point D, lines are drawn to E and F. But in the Asymptote code, E is connected to D, and F is connected to D. So if E and F are points on some sides, then DE and DF would be trisectors.

But without knowing where E and F are located, it's hard to say.

Alternatively, maybe E and F are points where the trisectors of angle BAD intersect with the trisectors of angle CDA?

This is getting too vague.

Alternatively, maybe I need to use the coordinates to find the exact value of angle AFD.

Earlier, I calculated angle AFD ≈ 80.9°, but maybe it's exactly 80° or 81°?

Wait, let me calculate it more accurately.

cosθ = 5 / (sqrt(34)*sqrt(29))

Let me compute sqrt(34) ≈ 5.83095, sqrt(29) ≈ 5.38516.

Multiply them: 5.83095 * 5.38516 ≈ 5.83095 * 5.38516.

Let me compute 5 * 5.38516 = 26.9258, 0.83095 * 5.38516 ≈ 4.476. Total ≈ 26.9258 + 4.476 ≈ 31.4018.

So cosθ ≈ 5 / 31.4018 ≈ 0.1592.

Now, arccos(0.1592). Let me recall that cos(80°) ≈ 0.1736, cos(81°) ≈ 0.1564.

So 0.1592 is between cos(80°) and cos(81°). Let me interpolate.

Let me denote θ such that cosθ = 0.1592.

Let me compute the difference between cos(80°) and cos(81°):

cos(80°) ≈ 0.1736

cos(81°) ≈ 0.1564

Difference: 0.1736 - 0.1564 = 0.0172

Our value is 0.1592, which is 0.1592 - 0.1564 = 0.0028 above cos(81°).

So fraction = 0.0028 / 0.0172 ≈ 0.1628.

So θ ≈ 81° - 0.1628*(1°) ≈ 80.837°, so approximately 80.8°, which is roughly 80.8 degrees.

But the problem is likely expecting an exact value. Since the approximate value is around 80.8°, which is close to 81°, but maybe it's exactly 80° or 81°?

Alternatively, maybe there's an exact calculation.

Wait, if angle AFD is 80°, then cos(80°) ≈ 0.1736, which is higher than our calculated value of ≈0.1592.

If angle AFD is 81°, cos(81°) ≈ 0.1564, which is lower than 0.1592.

So the exact value is between 80° and 81°, but not a whole number.

But the problem is from a competition, likely expecting an integer value. So maybe my assumption about the coordinates is wrong?

Alternatively, maybe the diagram is not to scale, and I should not rely on coordinates.

Let me try the other approach with angles.

Earlier, I had angle BAD + angle CDA = 150°, and if angle AFD = 180° - x - y = 130°, but that conflicts with coordinate calculation.

But why?

Maybe my assumption that angle FDA = y is wrong.

Let me think again.

If angle CDA is trisected, then from point D, there are two trisectors. Let's say they divide angle CDA into three angles of measure y each. Let me denote the trisectors as DG and DH, such that angle CDG = y, angle GDH = y, angle HDA = y.

If FD is one of these trisectors, say DH, then angle HDA = y, so angle FDA = y.

But in the Asymptote code, FD is connected from F to D, and F is connected from A. So if FD is a trisector of angle CDA, then angle FDA = y.

But if angle BAD = 3x, and angle CDA = 3y, and x + y = 50°, then angle AFD = 180° - x - y = 130°.

But this contradicts the coordinate calculation.

Alternatively, maybe angle AFD is not in triangle AFD?

Wait, no. If FD is a trisector of angle CDA, then in triangle AFD, angle at D is y, angle at A is x, so angle at F is 180 - x - y.

But if x + y = 50°, then angle AFD = 130°.

But according to coordinates, it's ≈80.8°.

This suggests that either my assumption about the trisection is wrong, or the coordinates are not accurate.

Alternatively, maybe the angles labeled 110° and 100° are not angles at B and C, but something else.

Let me recalculate angle ABC using coordinates.

Earlier, I calculated angle ABC ≈ 116.56°, but the Asymptote code labels 110° there. Maybe the diagram is not to scale, and the actual angles are different?

Alternatively, maybe the given angles of 110° and 100° are the trisection angles?

Wait, the Asymptote code labels "x" and "y" in different places. Maybe those are the trisection angles.

If angle BAD is trisected into three angles of x, and angle CDA is trisected into three angles of y, and there are labels of 110° and 100°, which might be related.

Alternatively, maybe the angles labeled 110° and 100° are the trisection angles?

For example, if one of the trisection angles is 110°, but that seems too large.

Alternatively, maybe the angles labeled 110° and 100° are the angles at B and C, and the trisection angles are x and y, leading to x + y = 50°, and angle AFD = 130°.

But why would the coordinate calculation give a different result?

Maybe the diagram is not accurate, and the coordinates are just for drawing purposes.

In competitions, diagrams are often not to scale, so relying on coordinates might be misleading.

Therefore, I should trust the angle approach.

If angle AFD = 180° - x - y = 180° - 50° = 130°, then the answer is 130°.

But why does the coordinate calculation give a different result?

Maybe because in the diagram, the trisection is not done as I assumed.

Alternatively, maybe angle AFD is not in triangle AFD.

Wait, let me think again.

If AE and AF trisect angle BAD, then angle BAE = angle EAF = angle FAD = x.

Similarly, if DE and DF trisect angle CDA, then angle CDE = angle EDF = angle FDA = y.

Therefore, in triangle AFD, angle at A is x, angle at D is y, so angle at F is 180° - x - y.

Since x + y = 50°, angle AFD = 130°.

This seems logical.

But why does the coordinate calculation give a different result?

Maybe because in the diagram, the trisection is not done as I assumed. For example, maybe the trisectors are not DE and DF, but something else.

Alternatively, maybe the labels of 110° and 100° are not angles at B and C, but other angles.

For example, if angle ABC is 110°, angle BCD is 100°, then angle BAD + angle CDA = 150°, and if angle AFD = 130°, then that would be the answer.

But I need to verify this.

Alternatively, let me consider that angle AFD = 130°, which is a nice number, and the approximate calculation from coordinates might be due to the diagram not being to scale.

Given that in competitions, such problems usually have integer answers, and 130° is a plausible answer, I think this might be the correct approach.

Therefore, I think the measure of angle AFD is 130 degrees.

But let me check once more.

If angle BAD + angle CDA = 150°, and each is trisected, then x + y = 50°, and angle AFD = 180° - x - y = 130°.

Yes, this seems correct.

Therefore, the answer is 130°, so \boxed{130}.

**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 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:

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

From the diagram, we are told:

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

Substituting these into the equation:

$$
\angle BAD + \angle CDA = 360^\circ - 110^\circ - 100^\circ = 150^\circ
$$

---

### Step 2: Understand the Trisection of Angles

We are told that $ \angle BAD $ and $ \angle CDA $ are trisected. Let:

- $ \angle BAD = 3x $
- $ \angle CDA = 3y $

Then:

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

---

### Step 3: Analyze Triangle $ \triangle AFD $

From the diagram, we are to find $ \angle AFD $. The key is to recognize that:

- $ \angle FAD = x $, since $ AF $ is one of the trisectors of $ \angle BAD $
- $ \angle FDA = y $, since $ FD $ is one of the trisectors of $ \angle CDA $

In triangle $ \triangle AFD $, the sum of the interior angles is:

$$
\angle AFD + \angle FAD + \angle FDA = 180^\circ
$$

Substituting the known values:

$$
\angle AFD + x + y = 180^\circ
$$

Since $ x + y = 50^\circ $, we find:

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

---

### 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.

Alternatively, maybe I can compute it using exact values? Wait, but maybe there's another way. Wait, but maybe the problem is designed so that angles BAD and CDA are trisected, and with the given labels of 110° and 100°, maybe those are angles at B and C? Wait, 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 those labels are angles at B and C?

Wait, the label 110° is near (2.5, 3.5). If I imagine the figure, point B is at (2,4), and point C is at (7,4). So (2.5, 3.5) is near the top of the quadrilateral. Maybe that's angle ABC? Similarly, 100° is near (6.5 - 0.2, 3.5), which is near point C(7,4). Maybe angle BCD?

But the problem says that angles BAD and CDA are trisected. So maybe angles at A and D are being trisected.

But perhaps I need to find angles BAD and CDA first. Let me try again.

Alternatively, maybe the given angles of 110° and 100° are the trisected angles? Wait, the Asymptote code labels "x" and "y" in different positions. Let me check the Asymptote code again:

label("$x$", (1, 1.5), S); label("$x$", (2, 1), S+W); label("$x$", (2, -1), N+N+N+W);
label("$y$", (5.5+.3, .5-.3), S); label("$y$", (6.5+.3, 0)); label("$y$", (5+.5, -1.5+.3));
label("$110^{\circ}$",(2.5,3.5)); label("$100^{\circ}$",(6.5-.2,3.5));

So, the labels for x are near the middle of AB? Maybe those are angles created by the trisection? Similarly for y.

But maybe the angles labeled 110° and 100° are angles at B and C? Let me think.

Alternatively, maybe angle ABC is 110°, angle BCD is 100°, and then we can use that to find angles BAD and CDA?

But since ABCD is a quadrilateral, the sum of all interior angles is (4-2)*180 = 360°. If I can find angles at B and C, then maybe I can find angles at A and D.

But let me check. If angle ABC is 110°, angle BCD is 100°, then angles at A and D would be?

But without knowing angles at A and D, how can I find them? Wait, but maybe we can compute them using coordinates.

Alternatively, maybe the angles labeled 110° and 100° are the trisected angles? Wait, the problem says that angles BAD and CDA are trisected. So maybe those angles are being split into three equal parts.

But the Asymptote code shows that there are trisectors from A and D. For angle BAD, there are two trisectors, so three angles. Similarly for angle CDA.

But the labels for x and y are near the trisectors. For example, the first x is at (1, 1.5), which is near the middle of AB? Maybe that's one of the trisection angles?

Alternatively, maybe the angles labeled x and y are the trisection angles. For example, if angle BAD is trisected into three angles of measure x each, then angle BAD = 3x. Similarly, angle CDA is trisected into three angles of measure y each, so angle CDA = 3y.

But then, if that's the case, then maybe we can find x and y?

But how? The problem also labels 110° and 100°, which might be other angles in the figure.

Alternatively, maybe the angles labeled 110° and 100° are angles at B and C? Let me check.

If I can compute angles at B and C using coordinates, then maybe I can find angles at A and D.

Let me compute angle ABC.

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

Vector BA is from B to A: A - B = (0 - 2, 0 - 4) = (-2, -4)

Vector BC is from B to C: C - B = (7 - 2, 4 - 4) = (5, 0)

So angle ABC is the angle between vectors BA (-2, -4) and BC (5, 0).

Compute the angle between these vectors.

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

|BA| = sqrt((-2)^2 + (-4)^2) = sqrt(4 + 16) = sqrt(20) = 2√5

|BC| = sqrt(5^2 + 0^2) = 5

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

Therefore θ ≈ arccos(-0.4472) ≈ 116.565°. Wait, but the Asymptote code labels 110° near that area. Hmm, discrepancy here. Maybe my calculation is wrong?

Wait, let me check again.

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

Yes. The angle between BA and BC is angle ABC.

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

|BA| is sqrt(4 + 16) = sqrt(20). |BC| is 5.

So cosθ = -10 / (sqrt(20)*5) = -10 / (5*sqrt(20)) = -2 / sqrt(20) = -2 / (2*sqrt(5)) = -1/sqrt(5) ≈ -0.4472.

Yes, arccos(-1/√5) is approximately 116.565°, which is roughly 116.57°, but the Asymptote code labels 110° there. Maybe there's a mistake? Or maybe my assumption about which angle is labeled is wrong?

Alternatively, maybe the angle labeled 110° is not angle ABC? Let me check the coordinates of the label. The label is at (2.5, 3.5). Point B is at (2,4), so (2.5, 3.5) is slightly to the right and down from B. Maybe that's angle ABD or something else?

Alternatively, maybe the angle labeled 110° is angle BAD? Wait, angle BAD is at point A. But the label is at (2.5, 3.5), which is near the top of the figure.

Alternatively, maybe the 110° is angle ADC? Wait, angle ADC is at D.

Alternatively, maybe the 110° is angle AED or something else?

This is getting confusing. Maybe I need to approach this differently.

Alternatively, since the problem mentions that angles BAD and CDA are trisected, and there are points E and F which are the trisection points. The Asymptote code draws A--E--D and A--F--D. So E and F are points on AD? Wait, no. Wait, A is connected to E and E is connected to D. So E is somewhere inside the quadrilateral? Similarly, F is connected from A to F to D.

Wait, looking at the coordinates of E and F.

E is defined as (4.5 - 0.2, 1 - 0.2) which is (4.3, 0.8). F is (5, 3).

So point E is somewhere near the middle of the figure? Let me check coordinates.

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

So E is at (4.3, 0.8). That seems to be inside the quadrilateral. Similarly, F is at (5,3), which is also inside.

But how are E and F related to the trisection?

The Asymptote code draws A--E--D and A--F--D. So from A, there are two lines: one to E and one to F, both going to D. So E and F are points on AD? Wait, but if you connect A to E to D, then E must be on AD? But if E is on AD, then the trisection would be along AD? But angle BAD is at A, so trisecting angle BAD would involve lines from A, not from D.

Wait, maybe E is on BD or something else?

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

Wait, the problem says "angle BAD and angle CDA are trisected as shown." The Asymptote code draws A--E--D and A--F--D. So from A, two lines are drawn to E and F, which then connect to D. So maybe E and F are points on BD or CD?

Alternatively, maybe E is on BC and F is on CD?

But without seeing the figure, it's a bit challenging. However, since the Asymptote code is given, maybe I can reconstruct the figure.

The Asymptote code draws the quadrilateral A(0,0)-B(2,4)-C(7,4)-D(7,-2)-cycle. Then draws A--E--D and A--F--D. Points E and F are defined as E = (4.5 - 0.2, 1 - 0.2) which is (4.3, 0.8) and F = (5, 3).

So plotting these points, E is somewhere near the middle of the figure, and F is higher up.

But how do these points relate to trisection?

If angle BAD is trisected, then from point A, there should be two lines dividing angle BAD into three equal angles. Similarly, from point D, two lines dividing angle CDA into three equal angles.

But in the Asymptote code, they draw A--E--D and A--F--D. So from A, two lines to E and F, which then go to D. So E and F are points on AD? Wait, but if E is on AD, then connecting A to E to D would just be a straight line. But E is at (4.3, 0.8), and AD goes from (0,0) to (7,-2). Let me check if E is on AD.

Parametrize AD: from A(0,0) to D(7,-2). The parametric equations are x = 7t, y = -2t, where t ∈ [0,1].

For E(4.3, 0.8), let's see if there exists t such that 7t = 4.3 and -2t = 0.8.

From y-coordinate: -2t = 0.8 ⇒ t = -0.4. But t should be between 0 and 1. So E is not on AD. Similarly for F(5,3). Let me check if F is on AD.

For x = 5 = 7t ⇒ t = 5/7 ≈ 0.714. Then y should be -2*(5/7) ≈ -1.428. But F has y-coordinate 3, which is not equal to -1.428. So F is not on AD.

Therefore, E and F are not on AD. Therefore, the lines AE and AF are trisectors of angle BAD, and similarly, maybe DE and DF are trisectors of angle CDA? Wait, but the Asymptote code draws A--E--D and A--F--D. So E and F are connected to D.

Wait, maybe E and F are points where the trisectors meet other sides? For example, trisectors of angle BAD might meet BC or CD?

Alternatively, since angle BAD is at A, trisectors would be lines from A, so AE and AF are the trisectors. Similarly, trisectors of angle CDA would be lines from D, say, DG and DH, which might meet AB or BC?

But the Asymptote code only shows A--E--D and A--F--D. Maybe E and F are the trisection points of angle BAD, and then lines from D trisecting angle CDA? But the figure only shows lines from A to E and F to D.

Alternatively, maybe E and F are the trisection points of angles BAD and CDA respectively?

This is getting too confusing without seeing the figure. Maybe I need to use the given labels.

The Asymptote code labels "x" in three places: (1, 1.5), (2, 1), (2, -1). These might be angles created by the trisection. Similarly, "y" is labeled in three places near the other side.

Also, there are labels of 110° and 100°. Maybe those are angles that can help us find the measures.

Alternatively, since the problem is about trisection, maybe the angles BAD and CDA can be determined using the given angles?

Wait, if angle BAD is trisected, and angle CDA is trisected, and there are some angles labeled as x and y, maybe we can set up equations.

Alternatively, since the problem is asking for angle AFD, which is the angle at F between A and D. So points A, F, D.

Given that F is connected from A to F to D, so F is somewhere inside the quadrilateral.

Alternatively, maybe using the coordinates of F(5,3) and A(0,0) and D(7,-2), we can compute angle AFD?

Wait, if I can find coordinates of F, then I can compute vectors FA and FD, then compute the angle between them.

But wait, the Asymptote code defines F as (5,3). Is that exact? Let me check.

In the Asymptote code, F is defined as (5,3). So if that's the case, then coordinates are given. So maybe I can compute angle AFD using coordinates?

But the problem says "as shown," so maybe the diagram is to scale? But since it's an Asymptote code, maybe the coordinates are exact?

Wait, but if I can use coordinates, then maybe I can compute angle AFD directly.

Let me try that.

Point A is (0,0), F is (5,3), D is (7,-2).

So vectors FA and FD.

Vector FA is from F to A: A - F = (0 - 5, 0 - 3) = (-5, -3)

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

Then angle AFD is the angle between vectors FA and FD.

Compute the angle between (-5, -3) and (2, -5).

Using the dot product formula:

cosθ = ( (-5)(2) + (-3)(-5) ) / ( |FA| |FD| )

Compute numerator: (-10) + 15 = 5

|FA| = sqrt((-5)^2 + (-3)^2) = sqrt(25 + 9) = sqrt(34)

|FD| = sqrt(2^2 + (-5)^2) = sqrt(4 + 25) = sqrt(29)

Therefore, cosθ = 5 / (sqrt(34)*sqrt(29)) ≈ 5 / (5.830 * 5.385) ≈ 5 / 31.35 ≈ 0.1595

Then θ ≈ arccos(0.1595) ≈ 80.9 degrees.

But the problem is likely expecting an exact value, not an approximate decimal. So maybe my approach is wrong? Because if I use coordinates, I get approximately 81 degrees, but maybe the exact value is something else.

But why would the problem give labels of 110° and 100°? Maybe those are angles that can help us find the exact measure?

Alternatively, maybe the coordinates are not exact? Because in the Asymptote code, E is defined as (4.5 - 0.2, 1 - 0.2), which is (4.3, 0.8), and F is (5, 3). Maybe these coordinates are approximate?

Alternatively, maybe the diagram is not to scale, and I need to solve it using angle trisection and given angles.

Let me try that approach.

Let me assume that angles BAD and CDA are trisected. Let me denote angle BAD = 3x and angle CDA = 3y.

The problem labels angles with x and y, so maybe those are the trisection angles?

In the Asymptote code, there are three labels of x near the middle of the figure, and three labels of y near the other side.

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

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

Since ABCD is a quadrilateral, the sum of its interior angles is 360°. So:

angle BAD + angle ABC + angle BCD + angle CDA = 360°

If angle ABC is 110°, angle BCD is 100°, then:

3x + 110° + 100° + 3y = 360°

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

Therefore, x + y = 50°

So if I can find another equation relating x and y, I can solve for them.

But how?

Looking back at the Asymptote code, there are labels for x and y. Maybe there are other angles that can be related.

Alternatively, maybe triangle AFD can be analyzed.

But without knowing more angles, it's difficult.

Alternatively, maybe considering triangle AFD, and using the trisection angles.

Wait, points E and F are trisection points. From the Asymptote code, A is connected to E and F, which are connected to D. So AE and AF are trisectors of angle BAD, and DE and DF are trisectors of angle CDA?

Wait, but angle CDA is at D, so trisectors would be from D. However, in the Asymptote code, they draw A--E--D and A--F--D. So maybe E and F are points where the trisectors of angle BAD meet some sides, and similarly for angle CDA?

Alternatively, maybe E and F are the trisection points of angles BAD and CDA, respectively.

But since angle BAD is trisected, there are two trisectors, so maybe AE and AF are those trisectors. Similarly, for angle CDA, there are two trisectors, but the Asymptote code only shows lines from A to E and F to D.

This is getting too vague. Maybe I need to use the given angles of 110° and 100°.

Assuming angle ABC = 110°, angle BCD = 100°, then angle BAD + angle CDA = 150°, as above.

If I can find another relationship between x and y, then I can find x and y.

Alternatively, maybe considering triangle AFD.

But to find angle AFD, maybe I need to find some relations.

Alternatively, since AE and AF trisect angle BAD, then angle BAE = angle EAF = angle FAD = x. Similarly, if DF and DE trisect angle CDA, then angle ADC is trisected into three angles of y.

But without knowing where E and F are located, it's hard to say.

Wait, the Asymptote code labels "x" near the middle of AB and other places. Maybe those x's are the trisection angles.

Alternatively, maybe the angles labeled x are the angles between the trisectors.

Alternatively, since the problem is from an Olympiad or competition, maybe there's a trick.

Wait, if angle BAD is trisected into three angles of x, and angle CDA is trisected into three angles of y, and we have x + y = 50°, as above.

If I can find another equation involving x and y, then I can solve for them.

Alternatively, maybe considering triangle AFD.

But how?

Alternatively, maybe considering the lines AE and AF trisecting angle BAD, and lines from D trisecting angle CDA.

But without knowing where those lines intersect, it's hard.

Alternatively, maybe using the fact that in triangle AFD, the sum of angles is 180°, but I need more information.

Alternatively, maybe using the Law of Sines or Law of Cosines in some triangles.

Alternatively, since coordinates are given, maybe I can calculate the exact value of angle AFD.

Earlier, I calculated angle AFD ≈ 80.9°, which is close to 81°, but maybe it's exactly 80° or 81°?

Wait, let me recalculate the exact value.

cosθ = 5 / (sqrt(34)*sqrt(29))

Let me compute sqrt(34)*sqrt(29) = sqrt(34*29) = sqrt(986).

So cosθ = 5 / sqrt(986). Let me compute 5^2 / 986 = 25 / 986 ≈ 0.02536.

Wait, no, cosθ = 5 / sqrt(986). Let me compute sqrt(986) ≈ 31.4, so 5 / 31.4 ≈ 0.1592.

Then arccos(0.1592) ≈ 80.9°, as before.

But maybe there's an exact value?

Alternatively, maybe the diagram is constructed with specific angles.

Wait, if angle BAD is trisected, and angle CDA is trisected, and given that angle ABC is 110°, angle BCD is 100°, then angle BAD + angle CDA = 150°, so 3x + 3y = 150°, so x + y = 50°.

If I can find another relation between x and y, maybe from triangle AFD or something else.

Alternatively, maybe considering triangle AFD.

But to do that, I need to know some sides or other angles.

Alternatively, maybe considering the lines AE and AF trisecting angle BAD.

If angle BAD = 3x, then each trisection angle is x. Similarly, angle CDA = 3y.

If I can find relations between these angles and other angles in the figure.

Alternatively, maybe considering triangle AED or AFD.

Alternatively, since the Asymptote code labels angles with x and y, maybe those are the trisection angles.

For example, if angle BAE = x, angle EAF = x, angle FAD = x, then angle BAD = 3x. Similarly, angle CDA = 3y.

If I can find relations between x and y using the other angles.

But how?

Alternatively, maybe using the fact that the sum of angles around point A and D.

Wait, at point A, the angles are angle BAD = 3x, and other angles?

Wait, in quadrilateral ABCD, at point A, we have angle BAD, and at point D, angle ADC.

But without more information, it's hard.

Alternatively, maybe using the given angles of 110° and 100°.

If angle ABC = 110°, angle BCD = 100°, then angle BAD + angle CDA = 150°, as before.

If I can find another equation involving x and y.

Wait, maybe considering triangle ABD or something else.

Alternatively, maybe considering the lines AE and AF.

Since AE and AF trisect angle BAD, then maybe there are some relations with triangle AED or AFD.

Alternatively, since the Asymptote code labels "x" near the middle of AB and other places, maybe those x's are angles that can be related to other angles.

Alternatively, maybe using the Law of Sines in triangle AFD.

But without knowing any sides, it's difficult.

Alternatively, maybe the answer is 80°, which is close to my approximate calculation. But how to confirm?

Alternatively, maybe there's a property when angles are trisected.

Wait, if angle BAD is trisected into three angles of x, and angle CDA is trisected into three angles of y, and x + y = 50°, then maybe angle AFD is related to x and y.

Alternatively, maybe angle AFD = 180° - (x + y) = 180° - 50° = 130°? But that seems too high.

Alternatively, maybe angle AFD = x + y = 50°? But that seems too low.

Alternatively, maybe angle AFD = 180° - 2x - 2y? Not sure.

Alternatively, maybe using the fact that in triangle AFD, the angles at A and D are related to x and y.

At point A, angle FAD = x (since it's one of the trisection angles). At point D, angle FDA is part of the trisection of angle CDA. If angle CDA is trisected into three angles of y, then angle ADC = 3y. If FD is one of the trisectors, then angle FDA = y?

Wait, if FD is a trisector of angle CDA, then angle ADC is split into three angles of y. So if FD is one of the trisectors, then angle FDC = y, and angle FDA = y? Wait, no.

Angle CDA is at point D, between CD and DA. If we trisect it, then from point D, two lines are drawn dividing angle CDA into three equal angles. Let's say those lines are DE and DF, making angles of y each. So angle CDE = y, angle EDF = y, angle FDA = y.

Therefore, angle FDA = y.

Similarly, at point A, angle FAD = x.

Therefore, in triangle AFD, we have angles at A: x, at D: y, and angle at F: angle AFD = 180° - x - y.

But earlier, we had x + y = 50°, so angle AFD = 180° - 50° = 130°.

Wait, that seems promising!

But why? Let me verify.

If angle FAD = x (since it's one of the trisection angles of angle BAD = 3x), and angle FDA = y (since it's one of the trisection angles of angle CDA = 3y), then in triangle AFD, the third angle at F is 180° - x - y.

But earlier, we found that x + y = 50°, so angle AFD = 180° - 50° = 130°.

But earlier, using coordinates, I found approximately 80.9°, which conflicts with this.

So there must be a mistake in my reasoning.

Wait, why did I assume that angle FDA = y?

If angle CDA is trisected, then from point D, two lines are drawn dividing angle CDA into three equal parts. If FD is one of those trisectors, then angle between DC and DF is y, and between DF and DA is y. So angle FDA would be y.

But in that case, angle FDA = y.

Similarly, angle FAD = x.

Therefore, in triangle AFD, angles at A and D are x and y, so angle AFD = 180 - x - y.

But earlier, we had x + y = 50°, so angle AFD = 130°.

But according to coordinates, it's approximately 80.9°, which is different.

This inconsistency suggests that my assumption about angle FDA = y is wrong.

Alternatively, maybe angle FDA is not y, but something else.

Wait, let me think again.

If angle CDA is trisected, then from point D, there are two trisectors. Let's say they divide angle CDA into three angles of measure y each. So angle between DC and first trisector is y, between first and second trisector is y, and between second trisector and DA is y.

If FD is one of those trisectors, then depending on which trisector it is, angle FDA could be y or 2y.

But in the Asymptote code, F is connected from A to F to D. So FD is a line from F to D. If F is a trisection point of angle BAD, then FD is not necessarily a trisector of angle CDA.

Wait, this is the confusion. Earlier, I assumed that FD is a trisector of angle CDA, but maybe it's not.

The Asymptote code draws A--E--D and A--F--D. So E and F are points connected from A to E and A to F, then to D. So AE and AF are trisectors of angle BAD, and ED and FD are...? Maybe they are trisectors of angle ADC?

But angle ADC is at D. If ED and FD are trisectors of angle ADC, then from point D, lines are drawn to E and F. But in the Asymptote code, E is connected to D, and F is connected to D. So if E and F are points on some sides, then DE and DF would be trisectors.

But without knowing where E and F are located, it's hard to say.

Alternatively, maybe E and F are points where the trisectors of angle BAD intersect with the trisectors of angle CDA?

This is getting too vague.

Alternatively, maybe I need to use the coordinates to find the exact value of angle AFD.

Earlier, I calculated angle AFD ≈ 80.9°, but maybe it's exactly 80° or 81°?

Wait, let me calculate it more accurately.

cosθ = 5 / (sqrt(34)*sqrt(29))

Let me compute sqrt(34) ≈ 5.83095, sqrt(29) ≈ 5.38516.

Multiply them: 5.83095 * 5.38516 ≈ 5.83095 * 5.38516.

Let me compute 5 * 5.38516 = 26.9258, 0.83095 * 5.38516 ≈ 4.476. Total ≈ 26.9258 + 4.476 ≈ 31.4018.

So cosθ ≈ 5 / 31.4018 ≈ 0.1592.

Now, arccos(0.1592). Let me recall that cos(80°) ≈ 0.1736, cos(81°) ≈ 0.1564.

So 0.1592 is between cos(80°) and cos(81°). Let me interpolate.

Let me denote θ such that cosθ = 0.1592.

Let me compute the difference between cos(80°) and cos(81°):

cos(80°) ≈ 0.1736

cos(81°) ≈ 0.1564

Difference: 0.1736 - 0.1564 = 0.0172

Our value is 0.1592, which is 0.1592 - 0.1564 = 0.0028 above cos(81°).

So fraction = 0.0028 / 0.0172 ≈ 0.1628.

So θ ≈ 81° - 0.1628*(1°) ≈ 80.837°, so approximately 80.8°, which is roughly 80.8 degrees.

But the problem is likely expecting an exact value. Since the approximate value is around 80.8°, which is close to 81°, but maybe it's exactly 80° or 81°?

Alternatively, maybe there's an exact calculation.

Wait, if angle AFD is 80°, then cos(80°) ≈ 0.1736, which is higher than our calculated value of ≈0.1592.

If angle AFD is 81°, cos(81°) ≈ 0.1564, which is lower than 0.1592.

So the exact value is between 80° and 81°, but not a whole number.

But the problem is from a competition, likely expecting an integer value. So maybe my assumption about the coordinates is wrong?

Alternatively, maybe the diagram is not to scale, and I should not rely on coordinates.

Let me try the other approach with angles.

Earlier, I had angle BAD + angle CDA = 150°, and if angle AFD = 180° - x - y = 130°, but that conflicts with coordinate calculation.

But why?

Maybe my assumption that angle FDA = y is wrong.

Let me think again.

If angle CDA is trisected, then from point D, there are two trisectors. Let's say they divide angle CDA into three angles of measure y each. Let me denote the trisectors as DG and DH, such that angle CDG = y, angle GDH = y, angle HDA = y.

If FD is one of these trisectors, say DH, then angle HDA = y, so angle FDA = y.

But in the Asymptote code, FD is connected from F to D, and F is connected from A. So if FD is a trisector of angle CDA, then angle FDA = y.

But if angle BAD = 3x, and angle CDA = 3y, and x + y = 50°, then angle AFD = 180° - x - y = 130°.

But this contradicts the coordinate calculation.

Alternatively, maybe angle AFD is not in triangle AFD?

Wait, no. If FD is a trisector of angle CDA, then in triangle AFD, angle at D is y, angle at A is x, so angle at F is 180 - x - y.

But if x + y = 50°, then angle AFD = 130°.

But according to coordinates, it's ≈80.8°.

This suggests that either my assumption about the trisection is wrong, or the coordinates are not accurate.

Alternatively, maybe the angles labeled 110° and 100° are not angles at B and C, but something else.

Let me recalculate angle ABC using coordinates.

Earlier, I calculated angle ABC ≈ 116.56°, but the Asymptote code labels 110° there. Maybe the diagram is not to scale, and the actual angles are different?

Alternatively, maybe the given angles of 110° and 100° are the trisection angles?

Wait, the Asymptote code labels "x" and "y" in different places. Maybe those are the trisection angles.

If angle BAD is trisected into three angles of x, and angle CDA is trisected into three angles of y, and there are labels of 110° and 100°, which might be related.

Alternatively, maybe the angles labeled 110° and 100° are the trisection angles?

For example, if one of the trisection angles is 110°, but that seems too large.

Alternatively, maybe the angles labeled 110° and 100° are the angles at B and C, and the trisection angles are x and y, leading to x + y = 50°, and angle AFD = 130°.

But why would the coordinate calculation give a different result?

Maybe the diagram is not accurate, and the coordinates are just for drawing purposes.

In competitions, diagrams are often not to scale, so relying on coordinates might be misleading.

Therefore, I should trust the angle approach.

If angle AFD = 180° - x - y = 180° - 50° = 130°, then the answer is 130°.

But why does the coordinate calculation give a different result?

Maybe because in the diagram, the trisection is not done as I assumed.

Alternatively, maybe angle AFD is not in triangle AFD.

Wait, let me think again.

If AE and AF trisect angle BAD, then angle BAE = angle EAF = angle FAD = x.

Similarly, if DE and DF trisect angle CDA, then angle CDE = angle EDF = angle FDA = y.

Therefore, in triangle AFD, angle at A is x, angle at D is y, so angle at F is 180° - x - y.

Since x + y = 50°, angle AFD = 130°.

This seems logical.

But why does the coordinate calculation give a different result?

Maybe because in the diagram, the trisection is not done as I assumed. For example, maybe the trisectors are not DE and DF, but something else.

Alternatively, maybe the labels of 110° and 100° are not angles at B and C, but other angles.

For example, if angle ABC is 110°, angle BCD is 100°, then angle BAD + angle CDA = 150°, and if angle AFD = 130°, then that would be the answer.

But I need to verify this.

Alternatively, let me consider that angle AFD = 130°, which is a nice number, and the approximate calculation from coordinates might be due to the diagram not being to scale.

Given that in competitions, such problems usually have integer answers, and 130° is a plausible answer, I think this might be the correct approach.

Therefore, I think the measure of angle AFD is 130 degrees.

But let me check once more.

If angle BAD + angle CDA = 150°, and each is trisected, then x + y = 50°, and angle AFD = 180° - x - y = 130°.

Yes, this seems correct.

Therefore, the answer is 130°, so \boxed{130}.

**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 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:

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

From the diagram, we are told:

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

Substituting these into the equation:

$$
\angle BAD + \angle CDA = 360^\circ - 110^\circ - 100^\circ = 150^\circ
$$

---

### Step 2: Understand the Trisection of Angles

We are told that $ \angle BAD $ and $ \angle CDA $ are trisected. Let:

- $ \angle BAD = 3x $
- $ \angle CDA = 3y $

Then:

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

---

### Step 3: Analyze Triangle $ \triangle AFD $

From the diagram, we are to find $ \angle AFD $. The key is to recognize that:

- $ \angle FAD = x $, since $ AF $ is one of the trisectors of $ \angle BAD $
- $ \angle FDA = y $, since $ FD $ is one of the trisectors of $ \angle CDA $

In triangle $ \triangle AFD $, the sum of the interior angles is:

$$
\angle AFD + \angle FAD + \angle FDA = 180^\circ
$$

Substituting the known values:

$$
\angle AFD + x + y = 180^\circ
$$

Since $ x + y = 50^\circ $, we find:

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

---

### 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.