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Croatian Mathematical Competitions

Geometry Difficulty 6.1 National Olympiad Prove it Croatia

The segment $\overline{AB}$ is a diameter of a circle with the centre $O$ . On the circle the point $C$ is given such that $OC$ is perpendicular to $AB$ . Let $P$ be a point on the shorter arc $\text{BC}$ . The lines $CP$ and $AB$ intersect at the point $Q$ , and the point $R$ is the intersection of the line $AP$ and the line through $Q$ perpendicular to the line $AB$ .
Prove that $|BQ| = |QR|$ . (Macedonia 2013)

Solution

The triangle $OCB$ is isosceles right triangle because $\overline{OB}$ and $\overline{OC}$ are both radii of the circle with the centre $O$ . Hence $\angle CBA = \angle CBO = 45^\circ$ .

Inscribed angles $\angle CPA$ and $\angle CBA$ over the chord $\overline{CA}$ are equal, so we have
$\angle QPR = \angle CPA = \angle CBA = 45^\circ.$

By Thales' Theorem the angle $\angle APB$ is right, so the angle $\angle BPR$ must be right as well. The quadrilateral $BQRP$ is cyclic because it has two opposite right angles ( $\angle RQB$ and $\angle BPR$ ), so the inscribed angles over the chord $QR$ are equal, which gives $\angle QBR = \angle QPR = 45^\circ$ .
Hence $\angle BRQ = 180^\circ - \angle RQB - \angle QBR = 45^\circ$ , so $\angle BRQ = 45^\circ = \angle QBR$ and we get $|BQ| = |QR|$ .

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