Quadrilateral $ABCD$ is inscribed in circle. Points $P$ and $Q$ lie respectively on rays $AB^{\rightarrow}$ and $AD^{\rightarrow}$ such that $AP = CD$, $AQ = BC$. Show that middle point of line segment $PQ$ lies on the line $AC$.
$AC$ is a median of $\triangle APQ$ is equivalent to:
$$\frac{\sin{\angle PAC}}{\sin{\angle QAC}}=\frac{AP}{AQ} \Leftrightarrow \frac{CD}{BC}=\frac{\sin{\angle BAC}}{\sin{\angle DAC}}$$Which is obvious by extended sine rule.