nice problem

I claim the concurrency point is the $F$-antipode $F'$. It is well-known that this is $\overline{CX} \cap \overline{IF}$. Let $P=\overline{EF} \cap \overline{DF'}$ and $Q=\overline{DF} \cap \overline{EF'}$. Then since $\angle PEQ=\angle PDQ=90^\circ$, $DEPQ$ is cyclic. Now, we have $$\angle EFD=\angle EFF'=90^\circ-\angle PF'E=90^\circ-\angle DFE=-90^\circ+\left(90^\circ-\frac{\angle A}{2}\right)+\left(90^\circ-\frac{\angle B}{2}\right)=\frac{\angle C}{2},$$so the center of $(DEPQ)$ lies on $(CDE)$ (and is on the same side of $\overline{DE}$ as $C$). On the other hand, it also lies on the perpendicular bisector of $\overline{DE}$, so it must be $C$ itself. This gives us $P=K$, and the conclusion follows. $\blacksquare$

IAmTheHazard wrote: I claim the concurrency point is the $F$-antipode $F'$. It is well-known that this is $\overline{CX} \cap \overline{IF}$. Let $P=\overline{EF} \cap \overline{DF'}$ and $Q=\overline{DF} \cap \overline{EF'}$. Then since $\angle PEQ=\angle PDQ=90^\circ$, $DEPQ$ is cyclic. Now, we have $$\angle EFD=\angle EFF'=90^\circ-\angle PF'E=90^\circ-\angle DFE=-90^\circ+\left(90^\circ-\frac{\angle A}{2}\right)+\left(90^\circ-\frac{\angle B}{2}\right)=\frac{\angle C}{2},$$so the center of $(DEPQ)$ lies on $(CDE)$ (and is on the same side of $\overline{DE}$ as $C$). On the other hand, it also lies on the perpendicular bisector of $\overline{DE}$, so it must be $C$ itself. This gives us $P=K$, and the conclusion follows. $\blacksquare$ can you explain why $angel PEQ$= $angel$ PDQ =90°