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Problem 956

AIME late

Geometry Difficulty 5.9 Prove it

1. In a convex quadrilateral $ABCD$ , $\angle DAB = \angle BCD = 90^{\circ}$ , $\angle ABC > \angle CDA$ , $Q$ and $R$ are points on segments $BC$ and $CD$ respectively, line $QR$ intersects $AB$ and $AD$ at points $P$ and $S$ respectively, and $PQ = RS$ . Let $M$ and $N$ be the midpoints of segments $BD$ and $QR$ respectively. Prove that points $A$ , $M$ , $N$ , and $C$ are concyclic.

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

1. As shown in Figure 1.

Since $N$ is also the midpoint of segment $PS$ , in the right triangles $\triangle P A S$ and $\triangle C Q R$ , we have respectively:
$\begin{array}{l} \angle A N P=2 \angle A S P, \\ \angle C N Q=2 \angle C R Q. \\ \text{Then } \angle A N C=\angle A N P+\angle C N Q \\ =2(\angle A S P+\angle C R Q) \\ =2(\angle R S D+\angle D R S) \\ =2 \angle A D C. \end{array}$

Similarly, in the right triangles $\triangle B A D$ and $\triangle B C D$ , we have:
$\angle A M C=2 \angle A D C.$

Therefore, $\angle A M C=\angle A N C$ .
Thus, points $A$ , $M$ , $N$ , and $C$ are concyclic.

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