Track / Stage 5 / 356 of 400 #956 of 1964

Problem 956

AIME late

Geometry Difficulty 5.9 Prove it

Inside triangle $A B C$ , a point $P$ is taken such that $\angle P A C=\angle P B C$ . Perpendiculars from point $P$ to sides $B C$ and $C A$ are dropped, intersecting at points $P M$ and $P K$ respectively. Let $D$ be the midpoint of side $A B$ . Prove that $D K=D M$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Prove the equality of triangles $K E D$ and $D F M$ , where $E$ and $F$ are the midpoints of $A P$ and $B P$ .

## Solution

Let $\angle P A C=\angle P B C=\alpha$ . If $E$ and $F$ are the midpoints of $A P$ and $B P$ respectively, then $\angle K E P=\angle M F P=2 \alpha$ . Since $D E$ and $D F$ are the midlines of triangle $A P B$ , $D E P F$ is a parallelogram.

$K E=E P=D F$ and $E D=F P=F M, \angle K E D=2 \alpha+\angle P E D=2 \alpha+\angle P F D=\angle M F D$ . Therefore, triangles $K E D$ and $D F M$ are equal by two sides and the included angle. Consequently, $D K=D M$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.