Proof of equal angles in a quadrilateral.

Question

points E and F are given on side BC of a convex quadrilateral ABCD (with E closer than F to B). Suppose angle EAB = angle CDF and angle FAE = angle FDE. Prove that angle CAF = angle EDB.

Answer 1

You are given that angle $FAE$ = angle $FDE$ so this means that $AEFD$ is a cyclic quadrilateral.

We will succeed in showing that angle $CAF$ = angle $EDB$ if we can show that angle $BAC$ = angle $BDC$, in otherwords if we can show that $ABCD$ is also a cyclic quadrilateral.

The picture here is, I hope, self-explanatory