Saudi Arabia Mathematical Competitions 2012

We shall prove that the circles $ADO$ and $BCO$ meet again at the incenter $I$ of the triangle $ABO$, so the line $IO$ is the radical axis of the circles $ADO$ and $BCO$. Noticing further that the lines $AP$ and $BC$ are the radical axes of the pairs of circles $(ABC,ADO)$ and $(ABC,BCO)$, respectively, it follows that the lines $AP$, $BC$ and $IO$ are concurrent (at point $Q$), and the conclusion follows.

To show that the point $I$ lies on the circle $ADO$, notice that $$\begin{array}{rl}\overline{AIO}& =90^{\circ }+\frac{1}{2}\overline{ABO}=90^{\circ }+\frac{1}{2}\overline{ABD}=90^{\circ }+\frac{1}{2}(180^{\circ }-2\overline{ADB})\\ & =180^{\circ }-\overline{ADB}=180^{\circ }-\overline{ADO}.\end{array}$$ Similarly, the point $I$ lies on the circle $BCO$, for $$\begin{array}{rl}\overline{BIO}& =90^{\circ }+\frac{1}{2}\overline{BAO}=90^{\circ }+\frac{1}{2}\overline{BAC}=90^{\circ }+\frac{1}{2}(180^{\circ }-2\overline{ACB})\\ & =180^{\circ }-\overline{ACB}=180^{\circ }-\overline{BCO}.\end{array}$$