Question:
"In triangle ABC, points E and D lie on AC and BC, respectively. Point F is inside the triangle such that $\angle CAF = \angle FAD$ and $\angle EBF$ and $\angle FBC$. Prove that $\angle AEB + \angle ADB = 2\angle AFB.$ "
Solution:
First, we label G at the intersection of AD and BE, we label H at the intersection of BE and AF, and we label I at the intersection of AD and BF. Let us say that angle CAF had a measure of x and angle CBF had a measure of y. Let us also say that angle EGA had a measure of z. Let's take the triangle EGA. We try to find angle AEB. It is $180-2x-z$ degrees. Since DGB is the vertical angle to EGA, we know that angle DGB has a measure of z. By doing the same thing we did on triangle EGA, we know that angle ADB is $180-2y-z$ degrees. Now we try to find angle AFB. Using the angles that we now know, we look at angle EHA. EHA is $180-(180-2x-z+x)=x+z$ degrees. Because of vertical angles, angle FHB is also $x+z$. Doing the same thing on the other side, angle FIA is $y+z$ degrees. We know that the measure of angle EGA+ the measure of angle EGD and that angle EGA is z degrees, EGD is $180-z$ degrees. We know that angle AFB is $360-(180-z+x+z+y+z)=180-x-y-z.$ Multiply that by two and we get $360-2x-2y-2z$ degrees. Adding the measures of AEB and BDA, we get $360-2x-2y-2z.$ Therefore, angle AEB+ angle ADB equals 2 * angle AFB.