Competition problem — Geometry — Maths Olympiad Prep

Library / /33 of 73

Geometry Difficulty 8.1 Shortlist Prove it Turkey

Let $D$ be the midpoint of the side $BC$ of a triangle $ABC$ and $AD$ intersect the circumcircle of $ABC$ for the second time at $E$ . Let $P$ be the point symmetric to the point $E$ with respect to the point $D$ and $Q$ be the point of intersection of the lines $CP$ and $AB$ . Prove that if $A, C, D, Q$ are concyclic, then the lines $BP$ and $AC$ are perpendicular.

Solution

Since $BD = DC$ and $PD = DE$ , we conclude that $BPCE$ is a parallelogram and hence $\angle PBC = \angle BCE = \angle BAE$ . Since the points $Q, A, C, D$ are concyclic, $\angle BAE = \angle QCB$ and therefore $\angle PBC = \angle QCB$ . Since $BD = DC$ , the lines $PD$ and $BC$ are perpendicular and $\angle PDC = \angle AQC = 90^\circ$ . Thus $AD$ and $CQ$ are altitudes of triangle $ABC$ and hence $BP$ is also an altitude. Done.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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