August 2018 Geometry#3

Let $H=AF \cap CD \cap BE$

Claim: $H$ is the incenter of $\Delta DEF$

Proof:

Note that $EHFC, DHFB$ and $EDBC$ are cyclic.

So $\angle EFH = \angle ECD = \angle DBE = \angle DFH$

Similarly for other angles, we obtain that $H \equiv I.$

Note that $\angle DBE = 27^{\circ}$ and $\angle DBC = 59^{\circ}$

So $\angle EBC = \angle CAF = \angle EDC = \angle CDF = 32^{\circ}$

$\therefore m\angle EDF = \boxed{64}.$