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Turkish Mathematical Olympiad

Geometry Difficulty 7.5 National Olympiad, round 2 Prove it Turkey

In an acute triangle $ABC$ points $D$ and $E$ are on the sides $[BC]$ and $[AC]$ , respectively, such that $BD$ and $CE$ are angle bisectors. Projections of $D$ onto $BC$ and $BA$ are $P$ and $Q$ , respectively, projections of $E$ onto $CA$ and $CB$ are $R$ and $S$ , respectively. Let $X$ be intersection of $AP$ and $CQ$ , $Y$ be the intersection of $AS$ and $BR$ , $Z$ be the intersection of $BX$ and $CY$ . Show that $AZ \perp BC$ .

Solution

Let $H$ be the projection of $B$ onto $AC$ . Since $B, P, Q, H, D$ lie on the semicircle of diameter $[BD]$ , we obtain
$\angle PHB = \angle PDB = 90^\circ - \frac{\hat{B}}{2} = \angle QDB = \angle QHB.$

We will show that if $BH \perp AC$ and $\angle PHB = \angle QHB$ , then $AP, CQ, BH$ are concurrent. Let the line passing through $B$ and parallel to $AC$ intersect the lines $PH$ and $QH$ at $K$ and $L$ , respectively. As $BH \perp AC$ and $\angle PHB = \angle QHB$ , we get $HB \perp KL$ and $\angle KHB = \angle LHB$ . Therefore, we have $|KB| = |LB|$ .

$KL \parallel AC$ and Thales theorem imply
$\frac{|KB|}{|CH|} = \frac{|BP|}{|PC|} \quad \text{and} \quad \frac{|LB|}{|AH|} = \frac{|BQ|}{|QA|}$
Since $|KB| = |LB|$ , we obtain $\frac{|AH|\,|PC|\,|BQ|}{|CH|\,|BP|\,|AQ|} = 1$ . Ceva theorem on triangle $ABC$ gives that $AP, CQ, BH$ are concurrent. Hence, $X \in BH$ and therefore, $BX \perp AC$ . Because of the symmetric structure we similarly get $CY \perp AB$ . Finally, the point $Z$ , which is the intersection of the lines $BX$ and $CY$ , is the orthocenter of the triangle $ABC$ and the result follows.

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