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Saudi Arabian IMO Booklet

Geometry Difficulty 7.7 National Olympiad, round 2 Prove it Saudi Arabia

Let $ABC$ be an acute-angled triangle. Point $P$ is such that $AP = AB$ and $PB \parallel AC$ . Point $Q$ is such that $AQ = AC$ and $CQ \parallel AB$ . Segments $CP$ and $BQ$ meet at point $X$ . Prove that the circumcenter of triangle $ABC$ lies on the circumcircle of triangle $PXQ$ .

Solution

Let $D$ be the vertex of parallelogram $ABDC$ . Then $APDC$ and $AQDB$ are isosceles trapezoids. Therefore the perpendicular bisectors to segments $PD$ and $QD$ coincide with the perpendicular bisectors to $AC$ and $AB$ respectively, the circumcenter $O$ of triangle $ABC$ is also the circumcenter of $DPQ$ and $\angle POQ = 2\angle A$ . Also since
$\angle XPD = \angle ADP, \angle XQD = \angle ADQ$
we obtain that $\angle PXQ = 2\angle A$ . Thus $O, P, Q, X$ are concyclic. $\square$

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