Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it

Let $ABC$ be a triangle. We denote $L, M, N$ as the midpoints of $[BC], [CA]$ , and $[AB]$ . Let $(d)$ be the tangent at $A$ to the circumcircle of $ABC$ . The line $(LM)$ intersects $(d)$ at $P$ , and the line $(LN)$ intersects $(d)$ at $Q$ . Show that $(CP)$ and $(BQ)$ are parallel.

Solution

Since $(A Q, A B) = (C A, C B) = (L Q, L B)$ , points $A, Q, B, L$ are concyclic, and similarly, points $A, P, C, M$ are concyclic. Therefore, $(C P, B C) = (C P, C L) = (A P, A L) = (A Q, A L) = (B Q, B L) = (B Q, B C)$ , so lines $(C P)$ and $(B Q)$ are parallel.

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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.