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Mongolian National Mathematical Olympiad

Geometry Difficulty 5.5 AIME, harder Prove it Mongolia

Let $ABCD$ be a cyclic quadrilateral with circumcenter $\omega$ , and $E$ be the intersection of the diagonals $AC$ and $BD$ . A line passing through $E$ intersects lines $AB$ , $BC$ at $P, Q$ , respectively. Let $R$ ( $R \neq D$ ) be the intersection point of $\omega$ and a circle that passes through $D, E$ and tangents the line $PQ$ at $E$ . Prove that $B, P, Q, R$ are cyclic.
(Proposed by B. Khoroldagva)

Solution

Let $\omega_1$ be the circle that passes through $D, E$ and tangents the line $PQ$ at $E$ . Since $PQ$ is tangent to the circle $\omega_1$ at $E$ we have $\angle EDR = \angle QER$ . So it follows from $\angle EDR = \angle BAR$ that $\angle PAR = \angle PER$ , i.e, $P, A, E, R$ are cyclic. Thus, since $\angle RBC = \angle RAE$ we get $\angle RPQ = \angle RBQ$ and so $P, B, Q, R$ are cyclic.

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