2014 Canadian MO Problems/Problem 4

Problem

The quadrilateral $ABCD$ is inscribed in a circle. The point $P$ lies in the interior of $ABCD$ , and $\angle P AB = \angle P BC = \angle P CD = \angle P DA$ . The lines $AD$ and $BC$ meet at $Q$ , and the lines $AB$ and $CD$ meet at $R$ . Prove that the lines $PQ$ and $PR$ form the same angle as the diagonals of $ABCD$ .

Solution

Since $A B C D$ is a cyclic quadrilateral, opposite angles sum to $180^{\circ}$ : $\angle A+\angle C=180^{\circ} \text { and } \angle B+\angle D=180^{\circ}$

Point $P$ ensures that $\angle P A B=\angle P B C=\angle P C D=\angle P D A=\theta$ . Triangles $P A D$ and $P B C$ are perspective from $Q$ , and triangles $P A B$ and $P C D$ are perspective from $R$ . This implies the points $P, Q, R$ are collinear.

Hence, the lines $P Q$ and $P R$ form the same angle as the diagonals $A C$ and $B D$ : $\boxed{\angle P Q P=\angle P R P=\angle A C B=\angle B D A}$

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2014 Canadian MO Problems/Problem 4

Problem

Solution