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Hong Kong competition problems, 2008

Geometry Difficulty 4.5 AIME Prove it Hong Kong

Let $H$ be an arbitrary point on the altitude $CP$ of an acute triangle $ABC$ . The lines $AH$ and $BH$ intersect $BC$ and $AC$ in $M$ and $N$ respectively. Show that $\angle NPC = \angle MPC$ .

Solution

(Blanchet's theorem) Suppose the line passing through $C$ and parallel to $AB$ meets $PN$ and $PM$ at $X$ and $Y$ respectively. Since $PC \perp XY$ , the result is the same as $CX = CY$ (as this implies $\triangle PCX \cong \triangle PCY$ ).

By Ceva's theorem, we have
$\frac{AP}{PB} \times \frac{BM}{MC} \times \frac{CN}{NA} = 1.$
Note that $\frac{BM}{MC} = \frac{PB}{YC}$ and $\frac{CN}{NA} = \frac{XC}{PA}$ by similar triangles. It follows that $CX = CY$ as desired.

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