Ukrainian National Mathematical Olympiad - Fourth Round

Let $M_1$ and $M_2$ be the middles of the diagonals (Fig. 26). Then $M_1M_2 \parallel BC$. Consider circumcircles of $\angle AQC$ and $\angle BPD$. Let them intersect $BD$ and $AC$ in points $X$ and $Y$. Since $BD$ contains the bisector of $\angle AQC$, then point $X$ is the middle of the bigger arc $AC$ of the circumcircle of $\angle AQC$. Then $X$ is on the perpendicular bisector of $AC$. In the same way, $Y$ is on the perpendicular bisector of $BD$. So $XM_1M_2Y$ is inscribed, as $\angle XM_1Y = \angle XM_2Y = 90^\circ$. Then $\angle M_1XM_2 = \angle M_1YM_2$. Also $\angle BXY = \angle CM_1M_2 = \angle M_1CB$, so $XBCY$ is inscribed. Hence, $\angle BXC = \angle BYC$.

$$\angle AQC = 180^\circ - 2\angle M_1XC = 180^\circ - 2(\angle M_1XM_2 + \angle M_2XC) = \\ = 180^\circ - 2(\angle M_1YM_2 + \angle M_1YB) = 180^\circ - 2\angle M_2YB = \angle BPD.$$

Alternative solution.

Denote on $BD$ point $Q'$ such that $\angle BQ'C = \angle BPC$. Then $BCQ'P$ is inscribed (Fig. 27). Hence $\angle Q'PC = \angle Q'BC = \angle Q'DA$, so $APQ'D$ is also inscribed. Then
$\angle CPD = \angle CPQ' + \angle Q'PD = \angle Q'DA + \angle Q'AD = \angle BQ'A.$
So $\angle BPD = \angle CQ'A$. Also $BD$ bisects $\angle CQ'A$. Let us show that $Q'=Q$. Assume these points are distinct. Because of the fact that