Autumn tournament

We claim the concurrency point is the $F$-antipode $F^{\prime }$. It is well-known that this is $\overline{CX}\cap \overline{IF}$. Let $P=\overline{EF}\cap \overline{D{F}^{\prime }}$ and $Q=\overline{DF}\cap \overline{E{F}^{\prime }}$. Then since $\angle PEQ=\angle PDQ=90^{\circ }$, $DEPQ$ is cyclic. Now, we have $$\begin{array}{rl}\angle EFD& =\angle EF{F}^{\prime }=90^{\circ }-\angle P{F}^{\prime }E=90^{\circ }-\angle DFE\\ & =-90^{\circ }+\left(90^{\circ }-\frac{\angle A}{2}\right)+\left(90^{\circ }-\frac{\angle B}{2}\right)=\frac{\angle C}{2},\end{array}$$ 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. $\square$