Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.8 AIME Prove it Belarus

Points $L$ and $H$ are marked on the sides $AB$ of an acute-angled triangle $ABC$ so that $CL$ is a bisector and $CH$ is an altitude. Let $P$ , $Q$ be the feet of the perpendiculars from $L$ to $AC$ and $BC$ respectively.
Prove that $AP \cdot BH = BQ \cdot AH$ . (I. Gorodnin)

Solution

We have
$\begin{align*} \frac{AP}{PH} &= \frac{AL \cos \angle A}{AC \cos \angle A} = \frac{AL}{AC} = [\text{since } CL \text{ is a bisectrix}] = \\ &= \frac{BL}{BC} = \frac{BL \cos \angle B}{BC \cos \angle B} = \frac{BQ}{BH}. \end{align*}$
It follows that $AP \cdot BH = BQ \cdot AH$ , as required.

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