IMO Hk TST, 1997

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

Given triangle $ABC$ , let $AL$ be the angle bisector of $\angle BAC$ with $L$ on $BC$ . Let the incircle of $\triangle ABC$ touch the sides $AB$ and $BC$ at points $P$ and $Q$ respectively. Let $X$ be the intersection point of the lines $AQ$ and $LP$ . Show that the lines $BX$ and $AL$ are perpendicular.

Solution

Let $I$ be the incentre of $\triangle ABC$ . Let $D$ be the projection of $B$ on $AL$ . Noting that $\triangle API \sim \triangle ADB$ and $\triangle LQI \sim \triangle LDB$ , we have $\frac{AP}{AD} = \frac{PI}{DB}$ and $\frac{LQ}{LD} = \frac{QI}{DB}$ . Therefore, we have
$\frac{AP}{PB} \times \frac{BQ}{QL} \times \frac{LD}{DA} = \frac{AP}{AD} \times \frac{LD}{LQ} = \frac{PI}{DB} \times \frac{DB}{QI} = 1.$
By Ceva's theorem, $AQ$ , $BD$ , $PL$ are concurrent at $X$ . Therefore, $BX \perp AL$ .

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