Track / Stage 6 / 109 of 400 #1109 of 2000

Problem 1109

National olympiad, first round

Geometry Difficulty 6.1 Prove it

Each side of the square $A B C D$ is divided into three equal parts, and the corresponding division points on opposite sides are connected by segments (see figure). Prove that $\angle A K M=\angle C D N$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Consider the auxiliary equal triangles.

## Solution

From the equality of right triangles $M L K$ and $D Q P$ (see the left figure), it follows that $\angle K M L = \angle P D Q$ . Since $L M K$ is the exterior angle of triangle $A K M$ , then $\angle A K M = \angle L M K - 45^{\circ}$ . Therefore, it remains to prove that $\angle P D N = 45^{\circ}$ .

From the equality of right triangles $D F P$ and $P E N$ , it follows that $D P = P N$ and $\angle D P N = 90^{\circ}$ . Consequently, $\angle C D N = \angle Q D P - \angle N D P = \angle Q D P - 45^{\circ} = \angle K M L - 45^{\circ} = \angle A K M$ .
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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.