Let $ABC$ be triangle with $A$-excenter $J$. The reflection of $J$ in $BC$ is $K$. The points $E$ and $F$ are on $BJ, CJ$ such that $\angle EAB=\angle CAF=90^{\circ}$. Prove that $\angle FKE+\angle FJE=180^{\circ}$.
Source: Baltic Way 2023/11
Tags: geometry
Let $ABC$ be triangle with $A$-excenter $J$. The reflection of $J$ in $BC$ is $K$. The points $E$ and $F$ are on $BJ, CJ$ such that $\angle EAB=\angle CAF=90^{\circ}$. Prove that $\angle FKE+\angle FJE=180^{\circ}$.