Baltic Way 2023 Shortlist

Let $JK$ intersect $BC$ at $X$. We'll prove a key claim:

Claim: $△BEK$ is similar to $△BAX$.

Proof. Note that $\angle EAB=90^{\circ }=\angle KXB$. Also, since $BJ$ bisects $\angle CBA$, we get $\angle ABE=\angle JBX=\angle XBK$. Hence $△EBA\sim △KBX$. From that, we see that the spiral similarity that sends the line segment $EA$ to $KX$ has center $B$. So the spiral similarity that sends the line segment $EK$ to $AX$ has center $B$. Thus $△BEK\sim BAX$. $\square$

In a similar manner, we get $△CFK$ is similar to $△CAX$.

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$$\begin{array}{rl}\angle FKE+\angle FJE& =\angle FKE+\angle BKC\\ & =360^{\circ }-\angle EKB-\angle CKF\\ & =360^{\circ }-\angle AXB-\angle CXA\\ & =360^{\circ }-180^{\circ }\\ & =180^{\circ }\end{array}$$ as desired.