Track / Stage 6 / 137 of 400 #1137 of 2000

Problem 1137

National olympiad, first round

Geometry Difficulty 6.2 Prove it

## Task A-1.3.

The length of $\overline{A B}$ is the diameter of a circle with center $O$ . On the circle, there is a point $C$ such that $O C$ is perpendicular to $A B$ . On the shorter arc $\overparen{B C}$ , a point $P$ is chosen. The lines $C P$ and $A B$ intersect at point $Q$ , and point $R$ is the intersection of the line $A P$ and the perpendicular through $Q$ to the line $A B$ .

Prove that $|B Q|=|Q R|$ .

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

## Solution.

Triangle $O C B$ is an isosceles right triangle because $\overline{O B}$ and $\overline{O C}$ are radii of the circle with center at point $O$ . Therefore, $\varangle C B A = \varangle C B O = 45^{\circ}$ .

The inscribed angles $\varangle C P A$ and $\varangle C B A$ subtended by the chord $\overline{C A}$ are equal, so

$\varangle Q P R = \varangle C P A = \varangle C B A = 45^{\circ}$

According to Thales' theorem, the angle $\varangle A P B$ is a right angle, so is $\varangle B P R$ .

Quadrilateral $B Q R P$ is cyclic because it has two right opposite angles ( $\varangle R Q B$ and $\varangle B P R$ ), so the inscribed angles over the chord $\overline{Q R}$ are equal, i.e., $\varangle Q B R = \varangle Q P R = 45^{\circ}$ .

Therefore, $\varangle B R Q = 180^{\circ} - \varangle R Q B - \varangle Q B R = 180^{\circ} - 90^{\circ} - 45^{\circ} = 45^{\circ}$ , so $\varangle B R Q = 45^{\circ} = \varangle Q B R$ , and it follows that $|B Q| = |Q R|$ .

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