Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.5 National olympiad Prove it Saudi Arabia

Circles $\omega_{1}$ and $\omega_{2}$ meet at $P$ and $Q$ . Segments $AC$ and $BD$ are chords of $\omega_{1}$ and $\omega_{2}$ respectively, such that segment $AB$ and ray $CD$ meet at $P$ . Ray $BD$ and segment $AC$ meet at $X$ . Point $Y$ lies on $\omega_{1}$ such that $PY \parallel BD$ . Point $Z$ lies on $\omega_{2}$ such that $PZ \parallel AC$ . Prove that points $Q, X, Y, Z$ are collinear.

Solution

Because quadrilateral $BPDQ$ is cyclic, we have $\angle PQD = \angle PBD$ . Because quadrilateral $ACQP$ is cyclic, we have $\angle PQC = 180^{\circ} - \angle CAP$ . We deduce that
$\begin{aligned} \angle DQC & = \angle PQC - \angle PQD = 180^{\circ} - \angle CAP - \angle PBD \\ & = 180^{\circ} - \angle XAB - \angle ABX = \angle BXA = \angle DXA. \end{aligned}$
This proves that quadrilateral $CQDX$ is cyclic and therefore
$\angle XQC = \angle XDC.$
Because $PA$ is parallel to $DX$ , we have $\angle YPC = \angle XDC$ . Because $CQPY$ is cyclic, we have $\angle YPC = \angle YQC$ . Therefore, $\angle YQC = \angle XQC$ , which means that points $Q, X, Y$ are collinear.

Because quadrilateral $BPQZ$ is cyclic, we have $\angle ZQB = \angle ZPB$ . Because lines $PZ$ and $AC$ are parallel, we have $\angle ZPB = \angle CAP$ .
Let $C'$ be the second intersection point of line $CQ$ with circle $\omega_{1}$ . Because quadrilateral $ACQP$ is cyclic, we have $\angle C'QP = \angle CAP$ . Therefore, $\angle ZQB = \angle C'QP$ . We deduce, by cyclicity of quadrilaterals $BPDQ$ and $CQDX$ , that
$\angle ZQC' = \angle BQP = \angle BDP = \angle XDC = \angle XQC,$
which means that points $Q, X, Z$ are collinear.

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