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Geometry Difficulty 7.6 National Olympiad, round 2 Prove it Hong Kong

Let $I$ be the incentre of a triangle $ABC$ with $AB > AC$ . A line through $I$ perpendicular to $AI$ cuts $BC$ produced at $P$ . Show that $IP^2 = BP \cdot CP$ .

Solution

We have
$\angle CIP = \angle CIA - 90^\circ = \angle CBI.$
This shows $IP$ is tangent to $(BCI)$ . Hence, $IP^2 = BP \cdot CP$ by considering the power of $P$ with respect to $(BCI)$ .

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