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Geometry Difficulty 8.2 Shortlist Prove it Baltic Way

Let $\triangle ABC$ be a triangle and let $J$ be the center of the excircle opposite to $A$ . The reflection of $J$ in $BC$ is $K$ . $E$ and $F$ are on $BJ$ and $CJ$ , respectively, such that $\angle EAB = \angle CAF = 90^\circ$ . Prove that $\angle FKE + \angle FJE = 180^\circ$ .

Solution

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.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.