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NMO Selection Tests for BMO and IMO

Geometry Difficulty 7.6 National Olympiad, round 2 Prove it Romania

Let $ABC$ be an acute-angled triangle and let $O$ be its circumcenter. The tangents of the circumcircle $ABC$ at vertices $B$ and $C$ meet at $P$ , the circle of radius $PB$ centered at $P$ meets the internal angle bisector of the angle $BAC$ at point $Q$ lying in the interior of the triangle $ABC$ , and the lines $OQ$ and $BC$ meet at $D$ . Finally, let $E$ and $F$ be the orthogonal projections of $Q$ on the lines $AC$ and $AB$ , respectively. Prove that the lines $AD, BE$ and $CF$ are concurrent.
Cosmin Pohoăţă

Solution

The line $AB$ and the circle of radius $PB$ centered at $P$ meet again at some point $R$ . Standard angle-chasing shows that the angle $BRC$ is the complement of the angle $BAC$ . Hence the lines $AC$ and $CR$ are perpendicular, so the lines $EQ$ and $CR$ are parallel, and the angles $CQE$ and $QCR$ are equal.

On the other hand, the line $OQD$ is the $Q$ -symmedian of the triangle $BCQ$ , so $BQ^2/CQ^2 = BD/CD$ . Consequently,
$1 = \frac{BD}{CD} \cdot \frac{CE}{BF} = \frac{BD}{CD} \cdot \frac{CE}{AE} \cdot \frac{AF}{BF},$
on account of $AQ$ being the internal angle bisector of the angle $BAC$ , so $AE = AF$ . The conclusion follows by Ceva's Theorem.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.