convex cyclic quadrilateral ABCD Points $E$ and $F$ are on side $BC$ of convex quadrilateral $ABCD$ (with $E$ closer than $F$ to $B$). It is known that $∠BAE = ∠CDF$ and $∠EAF = ∠FDE$. Prove that $∠FAC = ∠EDB$.
I have already deduced that ADFE is cyclic, and it suffices to prove ABCD is cyclic. However I am stuck at this point. I find it hard to use the angle condition $∠BAE = ∠CDF$.
 A: ADEF is cyclic, denote by $a$ its interior angle in $A$. Let $x$ be the common value of the two angles $\angle(BAE)$ and $\angle(FDC)$.
Then the angle in $A$ of $ABCD$ is $a+x$, and the exterior angle in $C$ of it is the exterior angle of the $\Delta CDF$ with angles $a$ (in $F$) and $x$ (in $D$), which is thus also $a+x$.
So $ABCD$ is cyclic.

A: Let $AB \cap DE = G$ and $AF \cap DC = H$. Then the angle condition gives $\angle GAH = \angle GDH$, so $GADH$ cyclic. The second angle condition gives $\angle EAF = \angle EDF$ so $AEFD$ cyclic.
Now, angle chasing, we get $\angle AFD = \angle AED = \angle EAG + \angle AGE$, and $\angle AFE = \angle HFC = \angle DCB - \angle CHF = \angle DCB - \angle DGA$.
Adding these angles gives $\angle FDC + \angle DCF = \angle DFE = \angle AFD + \angle AFE = \angle BAE + \angle DCB$, so $\angle BAE = \angle FDC$.
A: $ABCD$ is a cyclic quadrilateral [Theorem 1.9(EGMO)]
then angle $BAC$ = angle $CDB$ $[1]$
but angle $BAE$ = angle $CDF$;
and angle $EAF$ = angle $FDE$;
therefore angle $BAE$ + angle $EAF$ = angle $CDE$ + angle $FDE$;
which  is equivalent to angle $FAB$ = angle $CDE$.   $[2]$
Now, subracting 2 from 1 , we get --
angle $BAC$ - angle $FAB$ = angle $CDB$ - angle $CDE$
angle $FAC$ = angle $EDB$
$Q.E.D.$
