Balti Tee 2023 · Ülesanne 11

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

Claim: $\triangle BEK$ is similar to $\triangle BAX$.

Proof. Note that $\angle EAB = 90^\circ = \angle KXB$. Also, since $BJ$ bisects $\angle CBA$, we get $\angle ABE = \angle JBX = \angle XBK$. Hence $\triangle EBA \sim \triangle KBX$. 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 $\triangle BEK \sim BAX$. $\square$

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

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$$\begin{align*} \angle FKE + \angle FJE &= \angle FKE + \angle BKC \\ &= 360^\circ - \angle EKB - \angle CKF \\ &= 360^\circ - \angle AXB - \angle CXA \\ &= 360^\circ - 180^\circ \\ &= 180^\circ \end{align*}$$

as desired.