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Points $E$ and $F$ are on side $BC$ of convex quadrilateral $ABCD$ (with $E$ closer than $F$ to $B$). It is known that $\angle BAE =\angle CDF$ and $\angle EAF =\angle FDE$. Prove that $\angle FAC =\angle EDB$.

Simple angle chasing

Since, $\angle EAF = \angle FDE $ so $ADFE$ is cyclic.
So, $\angle FDA + \angle AEF = 180 $ degrees .....(1)

If we can prove that $ ADCB$ is a cyclic, then it'll be done.

Now, $\angle ABC + \angle ADC $= $(\angle FEA - \angle EAB) $ + $ (\angle FDA + \angle FDC)$ = $(\angle FDA + \angle AEF) +(\angle CDF - \angle BAE)$ = $180 $ degree [from (1) we get, $\angle FDA + \angle AEF = 180 $ degrees and $\angle BAE = \angle CDF$, as given]