Round 3

Geometry Difficulty 4.6 AIME Prove it Mongolia

Let $D$ be an interior point of a triangle $ABC$ such that $AD = DC$ . Let $M$ be the midpoint of $BC$ , $N$ be the foot of the perpendicular of $B$ to the line $DM$ and $L$ be the foot of the perpendicular of $N$ to the line $CD$ . Prove that the points $A, B, L, N$ lie on a circle.
(Proposed by Khulan Tumenbayar)

Solution

Choose a point $S$ on the ray $CD$ such that $SD = DC$ . Then $\angle SAN = 90^\circ = \angle SLN$ . It follows that the points $S, L, N, A$ lie on a circle.
Since $D$ is the midpoint of $SC$ and $M$ is the midpoint of $BC$ we have $DM \parallel SB$ . Therefore $\angle SBN = 90^\circ$ and hence the points $B, S, L, N, A$ lie on a circle.

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