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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 $\angle {FAC}=\angle{EDB}$.

AEFD is a cyclic quadrilateral . But we need to show $ABCD is a cyclic quadrilateral but how? Please help me. Thank you!

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As noted by you, we know that $AEFD$ is a cyclic quadrilateral, and so $\angle AEB = \angle ADF$.

We thus have that $$ \angle CBA + \angle ADC = \angle CBA + \angle ADF + \angle FDC = \angle CBA + \angle BAE + \angle AEB = 180^\circ. $$

Thus $ABCD$ is cyclic.

Diagram

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