Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME Prove it Ukraine

In triangle $ABC$ points $M$ and $N$ are the midpoints of $BC$ and $AC$ respectively. Inside this triangle point $P$ is chosen in such a way that $\angle BAP = \angle PCA = \angle MAC$ . Prove that $\angle PNA = \angle AMB$ .

Solution

It is easy to see that $\angle APC = 180^\circ - (\angle PCA + \angle PAC) = 180^\circ - \angle BAC = \angle ANM$ , since $MN \parallel AB$ as the centerline of the triangle (fig.14).

We know that $\angle MAC = \angle PCA$ , and so $\triangle MNA \sim \triangle APC$ . $N$ and $K$ are the midpoints of the respective sides in similar triangles, thus $\angle NKM = \angle PNA$ .

Now since $KN \parallel BC$ , we have $\angle NKM = \angle BMA = \angle PNA$ , and we are done.

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