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Geometry Difficulty 8.7 Shortlist Prove it Baltic Way

Let $I$ be the incenter of a triangle $ABC$ . Let $F$ and $G$ be the feet of the perpendiculars drawn from $A$ to the lines $BI$ and $CI$ , respectively. Rays $AF$ and $AG$ intersect the circumcircles of the triangles $CFI$ and $BGI$ second times at points $K$ and $L$ , respectively. Prove that line $AI$ bisects the segment $KL$ .

Figure 16

Solution

Since $\angle IFK = 90^\circ$ , then $IK$ is the diameter of the circumcircle of $CFI$ , hence also $\angle ICK = 90^\circ$ . Similarly is $IL$ the diameter of the circumcircle of $BGI$ and $\angle IBL = 90^\circ$ . Therefore are the lines $CK$ and $GL$ parallel, also $BL$ and $FK$ are parallel.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.