Prove sum of angles in problem involving bisectors in a given triangle

Question

Given: in $\triangle ABC$, $E$ and $D$ are points defined in sides $AC$ and $BC$, respectively. $AF$ bisects $\angle CAD$ and $BF$ bisects $\angle CBE$

Prove: $\angle AEB+\angle ADB=2\angle AFB$

I started with a draft of the problem for better visualization.

My strategy was to show that $\angle AFB$ can be constructed by the sum of 1/2 of angles $\angle AEB$ and $\angle ADB$. But after many attempts I'm not finding the answer.

Hints and solutions are welcomed. From a material with questions for math contests.

Answer 1

$$\angle{AEB}+\angle{ADB} = 2\pi - \angle{EAB} - \angle{DBA} - \angle{EBA} - \angle{DAB} = 2(\pi - \angle{FAD} - \angle{DAB} - \angle{EBA} - \angle{FBE}) = 2(\pi - \angle{FAB} - \angle{FBA}) = 2\angle{AFB}$$

Answer 2

Note that:

$$\angle CAB + \angle DAB = \angle FAB + \angle CAF + \angle FAB - \angle FAD = 2 \angle FAB$$

Similarly

$$\angle CBA + \angle EBA = 2 \angle FBA$$

Thus:

$$\angle AEB + \angle ADB = 180 - \angle CAB - \angle EBA + 180 - \angle CBA - \angle DAB$$

$$360 - 2\angle FBA - 2\angle FAB = 2\angle AFB$$

Hence the proof.