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Geometry Difficulty 7.7 National Olympiad, round 2 Prove it Hong Kong

$A$ and $B$ are fixed points on a plane and $L$ is a line passing through $A$ and not $B$ . $C$ is a variable point moving from $A$ toward infinity along a half-line of $L$ . The incircle of $\triangle ABC$ touches $BC$ at $D$ and $AC$ at $E$ . Show that line $DE$ passes through a fixed point.

Solution

Let $I$ be the incentre of $\triangle ABC$ , and let $P$ be the intersection point of $AI$ and $DE$ . We claim that $\angle APB = 90^\circ$ . Once this is proved, since the line $AP$ is fixed, the point $P$ is independent of $C$ , and so $P$ is the desired fixed point.

Note that $\angle BAI = \angle PAE$ and $\angle AIB = 90^\circ + \frac{C}{2} = \angle AEP$ . These imply $\triangle IAB \sim \triangle EAP$ . By spiral similarity, we have $\triangle PAB \sim \triangle EAI$ , and hence $\angle APB = \angle AEI = 90^\circ$ as desired.

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