37th Iranian Mathematical Olympiad

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



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