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Geometry Difficulty 4.5 AIME Prove it Slovenia

Let $ABCD$ be a square and $M$ the midpoint of the side $BC$ . Let $P$ be the orthogonal projection of point $C$ onto the line segment $DM$ . Prove that the triangle $DAP$ is an isosceles triangle with the top angle at $A$ .

Solution

Let $N$ be the midpoint of the side $CD$ , and let $R$ be the intersection point of the lines $DP$ and $AN$ . Then $\angle DRA = 180^\circ - \angle ADR - \angle NAD = 180^\circ - \angle ADR - \angle MDC = 180^\circ - 90^\circ = 90^\circ$ . The line $RN$ is thus parallel to the line $PC$ , and from $|DN| = |NC|$ we get $|DR| = |RP|$ . Since the triangles $DRA$ and $PRA$ coincide in two sides and the angle between them, they are congruent. We have $|DA| = |AP|$ .

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