Competition problem — Geometry — Olympiad Maths Prep

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Geometry Difficulty 6.6 National olympiad Prove it Iran

Consider an acute-angled triangle $\triangle ABC$ with $AB = AC$ and $\angle A > 60^\circ$ . Let $O$ be the circumcenter of $\triangle ABC$ . Point $P$ lies on the circumcircle of $\triangle BOC$ such that $BP \parallel AC$ , and point $K$ lies on segment $AP$ such that $BK = BC$ . Prove that line $CK$ bisects the arc $\widearc{BC}$ of circumcircle of $\triangle BOC$ .

Solution

Let $D$ be the second intersection point of circumcircle of $\triangle BOC$ and line $AC$ and $K'$ be the intersection point of lines $AP$ and $CM$ , where $M$ is the midpoint of arc $\widearc{BC}$ .

We have
$\begin{align*} \angle BCK' &= \frac{1}{2} \angle BOC = \angle A = \angle DAB, \\ \angle CDB &= 2\angle A \Rightarrow \angle ABD = \angle A \Rightarrow AD = BD. \end{align*}$
So we just need to prove that $\triangle BCK' \sim \triangle DAB$ to show that $K \equiv K'$ . Since
$\begin{align*} \angle ADP &= 180^\circ - \angle BPD = 180^\circ - \angle C, \\ \angle ACM &= \angle C + \frac{1}{2} \angle BOC = \angle C + \angle A, \end{align*}$
we get that $\angle ADP = \angle ACM$ . So, $PD \parallel CK'$ . Therefore $\frac{PD}{CK'} = \frac{AD}{AC} = \frac{AD}{AB}$ which is equivalent to $\frac{BC}{CK'} = \frac{AD}{AB}$ since $PB \parallel AC$ and $PD = BC$ . This completes the proof. ■

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