Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 8.3 Shortlist Prove it Romania

Let $ABC$ be a triangle, and let $M$ be a point on the side $AC$ , different from both $A$ and $C$ . The line through $M$ and parallel to $BC$ crosses the side $AB$ at $N$ . The segments $BM$ and $CN$ cross at $P$ , and the circles $BNP$ and $CMP$ meet again at $Q$ . Show that the angles $BAP$ and $CAQ$ are equal.

Solution

Invert from $A$ with radius $\sqrt{AB \cdot AM}$ , then reflect in the internal bisectrix of the angle $BAC$ , to obtain an involution $\varphi$ .
Since the lines $BC$ and $MN$ are parallel, $AB \cdot AM = AC \cdot AN$ , so $\varphi B = M$ and $\varphi C = N$ . The circles $ABM$ and $ACN$ are therefore the images under $\varphi$ of the lines $BM$ and $CN$ , respectively.

Since these two circles both pass through $Q$ , by Miquel's theorem, and $\varphi$ is an involution, $P = \varphi Q$ , so the lines $AP$ and $AQ$ are reflections of one another in the internal bisectrix of the angle $BAC$ . This ends the proof.

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