<Question>
In quadrilateral $ABCD$, angle $BAD$ and angle $CDA$ are trisected as shown. What is the degree measure of angle $AFD$?

[asy]
size(150);
pair A , B, C, D; A = (0,0); B = (2, 4); C = (7,4); D = (7, -2);
draw( (0,0)--(2,4) -- (7,4) -- (7, -2)-- cycle);
label("$A$", A, SW);
label("$B$", B, NW);
label("$C$", C, NE);
label("$D$", D, SE);
pair E, F;
E = (4.5-.2,1-.2); F = (5, 3);
draw(A--E--D); draw(A--F--D);
label("$E$", E, N); label("$F$", F, NW);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);
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));
[/asy]
</Question>

<think>
<Plan_1>
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. Let me try to visualize the diagram first. There's an Asymptote code provided, but since I can't actually see the image, I have to rely on the code and the labels.

From the Asymptote code, points are plotted as follows: A is at (0,0), B at (2,4), C at (7,4), D at (7,-2). So, ABCD is a quadrilateral with vertices at these coordinates. Then, E and F are points inside the quadrilateral. The Asymptote code draws lines from A to E to D, and A to F to D. The labels x, x, x near angles at A and D suggest that angles BAD and CDA are each trisected into three equal parts. There are also labels for 110 degrees and 100 degrees near points B and C.

The question is to find the measure of angle AFD.

First, let me recall that trisecting an angle means dividing it into three equal smaller angles. So if angles BAD and CDA are trisected, then each of those angles is split into three equal parts. The Asymptote code mentions labels x, x, x near those trisection points, so maybe each trisection angle is x degrees. Then, there are some other angles labeled y, y, y, and the angles at B and C are given as 110 and 100 degrees.

Wait, let me parse the Asymptote code's labels more carefully.

<prompt_1.1> Parse the angle labels from the Asymptote code to understand the trisection of angles BAD and CDA. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
The label "$x$" is placed at (1, 1.5) with direction South, which is probably near the trisection near point B. Then another x at (2,1) with SouthWest, and another x at (2,-1) with NorthNorthNorthWest. Similarly, the y labels are at (5.5+.3, .5-.3), (6.5+.3, 0), and (5.5, -1.5). Then there's a 110° label near (2.5,3.5) and a 100° label near (6.5,3.5). 

Hmm, perhaps the angles at A and D are trisected, creating angles labeled x and y. The 110° and 100° angles are at B and C. Wait, the Asymptote code draws the quadrilateral A-B-C-D-A. So, angle at B is 110°, angle at C is 100°, and angles at A and D are trisected. 

So, quadrilateral ABCD has angles at A and D trisected, with each trisection angle being x. Then, angles at B and C are given as 110° and 100°. The problem is to find angle AFD, which is formed by the intersection of AF and DF, where F is a point inside the quadrilateral connected to A and D.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
First, let's recall that the sum of the interior angles of a quadrilateral is 360 degrees. So, angles at A, B, C, D add up to 360°. Given that angles at B and C are 110° and 100°, their sum is 210°, so the sum of angles at A and D is 360° - 210° = 150°. Since angles at A and D are trisected, each trisection angle at A and D is x. But wait, if angles at A and D are trisected, then angle BAD is trisected into three angles of x each, and angle CDA is also trisected into three angles of x each? Wait, that might not make sense because angles at A and D are different.

<prompt_2.1> Calculate the trisection angles at points A and D using the sum of interior angles of the quadrilateral. </prompt_2.1>

Now, looking at the Asymptote code, points E and F are connected to A and D, creating triangles AED and AFD. The labels x, x, x near the trisection might indicate that each trisection angle is x. Since angle BAD is trisected, each of those angles is x, so angle BAD = 3x. Similarly, angle CDA is trisected into three angles of x each, so angle CDA = 3x. Wait, but earlier we found angle BAD + angle CDA = 150°, so 3x + 3x = 150°, so 6x = 150°, which gives x = 25°. Therefore, each trisection angle is 25 degrees.

Wait, but angle BAD is trisected, so angle BAD is divided into three equal parts of 25° each. Similarly, angle CDA is also trisected into three parts of 25° each. That makes sense. So, angles at A and D are each 75°, since 3*25° = 75°. Then, angles at B and C are 110° and 100°, which sum up with 75 + 75 to give 360.
</Plan_2>

<Execution_2>
<execution_2.1>
Wait, actually, angle BAD is at vertex A, and angle CDA is at vertex D. So, angle BAD is trisected, meaning angle BAD is divided into three equal angles, each of measure x. Similarly, angle CDA is trisected into three equal angles, each of measure x. Wait, but if both angles at A and D are trisected into x's, then angle BAD = 3x and angle CDA = 3x. Then, as we found, 3x + 3x + 110 + 100 = 360. So, 6x = 150, so x = 25.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
So, each trisection angle is 25 degrees. That might be important for finding angle AFD.

Looking at the Asymptote code, point E is connected to A and D, and point F is also connected to A and D. The labels for x are near the trisection marks, so maybe each x is 25 degrees. The Asymptote code also labels angles 110° and 100°, which are probably angles at points B and C.

<prompt_3.1> Analyze the structure of the quadrilateral and identify the triangles formed by points A, F, and D. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
So, angle at B is 110°, angle at C is 100°, angle at A is 75°, angle at D is 75°, summing to 360.

Now, the problem is to find angle AFD. Point F is the intersection of lines AF and DF. Wait, no, in the Asymptote code, it's drawn as A connected to F connected to D, so AF and FD form angle AFD at point F. So, angle AFD is the angle at point F between points A and D. To find this angle, we might need to consider triangle AFD and use some trigonometric laws, like the Law of Sines or Law of Cosines, if we can find the necessary sides or other angles.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Alternatively, maybe there's a way to use the properties of the trisection and the given angles to find angle AFD.

First, let me try to figure out the coordinates of points E and F. Wait, the Asymptote code gives coordinates: A(0,0), B(2,4), C(7,4), D(7,-2). Then, E is at (4.5 - 0.2, 1 - 0.2) = (4.3, 0.8), and F is at (5,3). But wait, in the Asymptote code, the drawing is A-B-C-D-A. So, the quadrilateral is A(0,0) connected to B(2,4), then to C(7,4), then to D(7,-2), then back to A(0,0). Then, E is near (4.3, 0.8), which is somewhere between C(7,4) and D(7,-2), but closer to D. Wait, but E is labeled near the trisection near A and B. Hmm, maybe the coordinates are approximate.

<prompt_4.1> Consider the possibility of using coordinates to solve for angle AFD and verify with the Law of Sines. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Alternatively, maybe the figure is not to scale, and we need to solve it using geometric properties rather than coordinates.

Alternatively, if I can compute coordinates based on the given angles, maybe I can compute the exact position of F and then compute angle AFD. But this might be complicated. Alternatively, maybe there's a way to use angle chasing.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Since angles at A and D are each 75°, and they are trisected into 25° each. So, at point A, angle BAD is 75°, trisected into three 25° angles. Therefore, the lines from A to E and A to F trisect angle BAD. Similarly, at point D, angle CDA is 75°, trisected into three 25° angles, so lines from D to E and D to F trisect angle CDA.

Wait, but in the Asymptote code, E is connected to A and D, and F is connected to A and D. So, lines AE and DF are drawn? Wait, no, the code says draw(A--E--D); draw(A--F--D); so, A connected to E connected to D, and A connected to F connected to D. So, triangles AED and AFD are drawn.

<prompt_5.1> Use angle chasing and properties of triangles to find angle AFD. </prompt_5.1>

Alternatively, maybe we can use the Law of Sines in triangle AFD.

<prompt_5.2> Apply the Law of Sines in triangle AFD to find angle AFD. </prompt_5.2>

But wait, maybe there's a better way. Let me consider the complete quadrilateral formed by points A, B, C, D, E, F. Alternatively, maybe using the concept of triangle angle sums.

<prompt_5.3> Explore alternative methods, such as considering triangle AFD and the sum of angles, to find angle AFD. </prompt_5.3>

But let me verify this with another approach. Let's consider triangle AFD.

<prompt_5.4> Verify the measure of angle AFD using the Law of Sines in triangle AFD. </prompt_5.4>

Alternatively, perhaps using the Law of Sines in triangle AFD. If we can find the ratio of sides AF and FD, but since we don't have lengths, maybe it's tricky. Alternatively, maybe using the Law of Sines in triangles involving E.

<prompt_5.5> Explore the use of the Law of Sines in triangles involving points E and F to verify angle AFD. </prompt_5.5>

Alternatively, maybe there's a way to use the given angles at B and C. Since angle at B is 110°, and angle at C is 100°, perhaps considering triangle ABC or BCD.

<prompt_5.6> Use the given angles at B and C to verify the measure of angle AFD. </prompt_5.6>

Alternatively, maybe considering the full quadrilateral and the sum of angles around point F. However, without knowing other angles at F, this might not be straightforward.

<prompt_5.7> Consider the sum of angles around point F and verify the measure of angle AFD. </prompt_5.7>

Alternatively, let me check with another method. Suppose I use coordinates. Let's assign coordinates to the points and compute the angles.

<prompt_5.8> Assign coordinates to points and use them to verify the measure of angle AFD. </prompt_5.8>

Alternatively, maybe using vector methods. Let me try that.

<prompt_5.9> Use vector methods to verify the measure of angle AFD. </prompt_5.9>
</Plan_5>

<Execution_5>
<execution_5.1>
So, perhaps triangles AED and AFD are involved.

Wait, but point E is on BD, maybe? Wait, in the Asymptote code, E is plotted near (4.5 - 0.2, 1 - 0.2), which is approximately (4.3, 0.8). Given that B is at (2,4), C is at (7,4), D is at (7,-2), so BD is from (2,4) to (7,-2). The line BD can be parametrized. Let me compute the coordinates of E.

But maybe coordinates are complicating things. Let's try to solve this using pure geometry first.
</execution_5.1>

<execution_5.2>
For that, I need to know some side lengths or other angles in triangle AFD. Alternatively, maybe triangle AFD is part of a larger system of triangles where we can apply the Law of Sines or Cosines multiple times.

Alternatively, since angles at A and D are trisected, maybe triangles ABE, EBF, FBD, etc., can be considered. Wait, but without knowing the exact positions of E and F, this might be complicated. Alternatively, maybe considering that lines AE and DF trisect angles at A and D, so angles between BA and AE is 25°, between AE and AF is another 25°, and between AF and AD is another 25°. Similarly at D, angles between CD and DF, DF and DE, and ED and DA are each 25°.

Wait, perhaps if I can model the angles around points A and D, and use that to find the necessary angles in triangle AFD.
</execution_5.2>

<execution_5.3>
Alternatively, maybe considering triangle AFD and using the Law of Sines or Cosines. But to do that, we need to know some side lengths or other angles. Alternatively, maybe considering the angles around point F.

Wait, point F is inside the quadrilateral, connected to A and D. So, angle AFD is formed by the intersection of AF and DF. To find angle AFD, perhaps we can consider the angles at A and D and how they relate to the angles at F.
</execution_5.3>

<execution_5.4>
To find angle AFD, we need to know the other angles in triangle AFD. Let's see. If we can find angles at A and D in triangle AFD, then we can find angle AFD. 

At point A, angle BAD is 75°, which is trisected into three 25° angles. So, angle BAE = 25°, angle EAF = 25°, and angle FAD = 25°. Similarly, at point D, angle CDA is 75°, trisected into three 25° angles. So, angle ED C = 25°, angle FDE = 25°, and angle FDA = 25°. Wait, but how exactly are the trisection lines arranged? 

If AE and AF trisect angle BAD, then starting from AB, the first trisection would be AE, then AF, then AD. Similarly, at D, if DF and DE trisect angle CDA, then starting from DC, the first trisection would be DF, then DE, then DA. Wait, but the labels in the Asymptote code suggest that E is near the trisection near B, and F is somewhere else. 

Alternatively, perhaps the trisection is such that angle BAE = 25°, angle EAF = 25°, angle FAD = 25°, and at D, angle FDE = 25°, angle EDF = 25°, angle CDF = 25°. But I need to confirm the exact configuration.
</execution_5.4>

<execution_5.5>
Alternatively, since angles at A and D are trisected, maybe we can find some relationships between the triangles.

Alternatively, let's consider the lines from A and D. Since angle BAD is 75°, trisected into three 25° angles, so angle BAE = 25°, angle EAF = 25°, angle FAD = 25°. Similarly, at point D, angle CDA is 75°, trisected into three 25° angles. So, angle CDF = 25°, angle FDE = 25°, angle EDA = 25°. Wait, but the labels in the Asymptote code have y's. Wait, in the Asymptote code, there are labels for y at (5.5+.3, .5-.3), (6.5+.3, 0), and (5.5+.3, -1.5+.3). So maybe those trisection angles at D are labeled y, but since we found that each trisection angle is 25°, which is x. Wait, maybe the labels are mixed. Wait, in the Asymptote code, angles labeled x are near the trisection near A and B, and angles labeled y are near the trisection near D. But since angles at A and D are both 75°, trisected into 25°, so both x and y would be 25°. But maybe in the problem statement, they are using x for A's trisection and y for D's trisection. But since both are 25°, maybe x = y =25°. Then, angles at B and C are 110° and 100°, which are given.

Alternatively, maybe the problem is similar to some standard configurations. For example, in a quadrilateral with two opposite angles trisected, and two other angles given, find a specific angle formed by the trisectors.
</execution_5.5>

<execution_5.6>
Wait, but ABCD is a quadrilateral, not necessarily a triangle. Alternatively, maybe considering triangle AFD and the angles at A and D. At point A, angle BAD is 75°, which is trisected into three 25° angles. So, if AF is one of the trisection lines, then angle BAF is 25°, angle FAD is 25°, and angle DAF is 25°, but wait, angle BAD is 75°, so three trisection angles of 25° each. So, if AF is the first trisection line from AB, then angle BAF = 25°, and angle FAD = 25°, but wait, angle BAD is 75°, so maybe AF and FD are the trisection lines. Wait, but FD is connected to D, so maybe FD is one of the trisection lines at D. Hmm, maybe this is getting confusing.

Alternatively, maybe considering triangle AFD. In triangle AFD, we know angles at A and D. Wait, angle at A in triangle AFD would be angle FAD. Since angle BAD is 75°, and AF trisects angle BAD, then angle FAD is 25°. Similarly, angle at D in triangle AFD would be angle FDA. Since angle CDA is 75°, and DF trisects angle CDA, then angle FDA is 25°. Therefore, triangle AFD has two angles of 25°, so the third angle at F would be 180° - 25° - 25° = 130°. Therefore, angle AFD is 130°.
</execution_5.6>

<execution_5.7>
Wait, another approach: since angles at A and D are each 75°, trisected into 25°, and angles at B and C are 110° and 100°, perhaps we can find the angles at E and F using some properties.

Alternatively, since lines AE and DF trisect angles at A and D, maybe triangles AED and AFD have some properties. Wait, but I need more information.
</execution_5.7>

<execution_5.8>
Given points A(0,0), B(2,4), C(7,4), D(7,-2). Let's compute vectors and angles.

First, angle at B is 110°, which is the angle between BA and BC. Vector BA is A - B = (-2, -4), and vector BC is C - B = (5, 0). The angle between BA and BC can be computed using the dot product:

cos(theta) = (BA . BC) / (|BA| |BC|)

BA . BC = (-2)(5) + (-4)(0) = -10

|BA| = sqrt(4 + 16) = sqrt(20) = 2*sqrt(5)

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

So, cos(theta) = -10 / (2*sqrt(5)*5) = -10 / (10*sqrt(5)) = -1/sqrt(5)

Therefore, theta = arccos(-1/sqrt(5)) ≈ 116.565°, but the problem states it's 110°. Hmm, discrepancy here. Maybe the coordinates are not to scale? Wait, the problem might not be using the exact coordinates but a diagram where those angles are given as 110° and 100°, so perhaps the coordinates are just for drawing purposes and not exact. Therefore, using coordinates might not be accurate here.
</execution_5.8>

<execution_5.9>
First, assign coordinates to points. Let's set point A at (0,0). Then, since angle BAD is 75°, and it's trisected into 25° each. So, from point A, the lines AE and AF make angles of 25° and 50° with respect to AB. Wait, but we need to know the direction of AB. Wait, in the Asymptote code, AB is from (0,0) to (2,4), so AB has a slope of 2. Then, angle BAD is the angle between BA and AD. Wait, but in the Asymptote code, the quadrilateral is A(0,0) connected to B(2,4), then to C(7,4), then to D(7,-2), then back to A. So, angle at A is between AB and AD. Let's compute the angle at A.

Coordinates: A(0,0), B(2,4), D(7,-2). So vector AB is (2,4), vector AD is (7,-2). The angle between AB and AD can be found using the dot product.

Dot product of AB and AD vectors: (2)(7) + (4)(-2) = 14 - 8 = 6.

Magnitude of AB: sqrt(2² + 4²) = sqrt(4 + 16) = sqrt(20) = 2*sqrt(5).

Magnitude of AD: sqrt(7² + (-2)²) = sqrt(49 + 4) = sqrt(53).

Therefore, cos(theta) = 6 / (2*sqrt(5)*sqrt(53)) ) = 6 / (2*sqrt(265)) = 3 / sqrt(265) ≈ 3 / 16.28 ≈ 0.184.

So theta ≈ arccos(0.184) ≈ 80 degrees. Wait, but earlier we thought angle at A was 75 degrees. There's a discrepancy here. Hmm. So perhaps the coordinates given in the Asymptote code do not align with the problem statement? Because according to the coordinates, angle at A is approximately 80 degrees, but according to the problem statement, angle at A is 75 degrees, trisected into 25 degrees each. Therefore, the coordinates might be illustrative but not exact. Therefore, using coordinates might not be accurate here. So perhaps coordinate geometry is not the way to go here.
</execution_5.9>
</Execution_5>

<Plan_6>
Based on execution_5.6:
Wait, but is this correct? Let me check again.

If angle BAD is 75°, and it's trisected into three angles of 25° each, then angle BAF would be 25°, angle FAD would be 25°, and angle DAF would be 25°, but since BAD is 75°, that's three times 25°, which adds up. Similarly, angle CDA is 75°, trisected into three angles of 25°, so angle FDA would be 25°, angle EDF would be 25°, and angle EDC would be 25°, but again, angle CDA is 75°, so three times 25°. 

Therefore, in triangle AFD, angles at A and D are each 25°, so angle AFD is 180° - 25° - 25° = 130°.
That seems straightforward.
</Plan_6>

<Execution_6>
No prompt to execute.
</Execution_6>

<Final_answer>
Therefore, the degree measure of angle $AFD$ is $\boxed{130}$.
</Final_answer>
</think>