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Geometry Difficulty 4.5 AIME Prove it JBMO

Problem:

Let the inscribed circle of the triangle $\triangle ABC$ touch side $BC$ at $M$ , side $CA$ at $N$ and side $AB$ at $P$ . Let $D$ be a point from $[NP]$ such that $\frac{DP}{DN}=\frac{BD}{CD}$ . Show that $DM \perp PN$ .

Solution

Solution:

From $AP = AN$ it follows that $\angle ANP = \angle APN$ or $\angle NPB = \angle PNC$ (both obtuse). Hence the triangles $BDP$ and $CND$ are similar (SSA) and $\angle CDN = \angle BDP$ and $\frac{CD}{BD} = \frac{CN}{BP} = \frac{CM}{BM}$ . So $DM$ is a bisector of the angle $BDC$ , from where $NP \perp MD$ .

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