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 $∠BAE = ∠CDF$.
Best Answer
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.