Belarus — Selection and Training Session

We refer to the following well-known fact: in any cyclic quadrilateral all perpendiculars drawn through the midpoints of the sides to the opposite sides meet at the same point $U$. It follows that the altitudes $DD_1$ and $CC_1$ of the triangle $ADB$ and $BCP$, respectively, are symmetric with respect to $U$. The same is true for the altitudes $AA_1$ and $BB_1$. Therefore the intersection points $DD_1 \cap AA_1$ and $CC_1 \cap BB_1$ are symmetric with respect to $U$. That is, the orthocenters $H_{ADP}$ and $H_{BCP}$ are symmetric. Similarly, the orthocenters $H_{AQB}$ and $H_{DQC}$ are symmetric with respect to $U$. So the segments $H_{ADP}H_{AQB}$ and $H_{CQD}H_{BCP}$ are symmetric, so they have the same length.