Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.0 National olympiad, round 2 Prove it

Let $A B C$ be a triangle and let $D$ be a point on the segment $B C, D \neq B$ and $D \neq C$ . The circle $A B D$ meets the segment $A C$ again at an interior point $E$ . The circle $A C D$ meets the segment $A B$ again at an interior point $F$ . Let $A^{\prime}$ be the reflection of $A$ in the line $B C$ . The lines $A^{\prime} C$ and $D E$ meet at $P$ , and the lines $A^{\prime} B$ and $D F$ meet at $Q$ . Prove that the lines $A D, B P$ and $C Q$ are concurrent (or all parallel).

Solution

. (Ilya Bogdanov) Let $\sigma$ denote reflection in the line $B C$ . Since $\angle B D F=\angle B A C=$ $\angle C D E$ , by concyclicity, the lines $D E$ and $D F$ are images of one another under $\sigma$ , so the lines $A C$ and $D F$ meet at $P^{\prime}=\sigma(P)$ , and the lines $A B$ and $D E$ meet at $Q^{\prime}=\sigma(Q)$ . Consequently, the lines $P Q$ and $P^{\prime} Q^{\prime}=\sigma(P Q)$ meet at some (possibly ideal) point $R$ on the line $B C$ .

Since the pairs of lines $(C A, Q D),(A B, D P),(B C, P Q)$ meet at three collinear points, namely $P^{\prime}, Q^{\prime}, R$ respectively, the triangles $A B C$ and $D P Q$ are perspective, i.e., the lines $A D, B P, C Q$ are concurrent, by the Desargues theorem.
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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.