62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour

By construction $\Delta ADK=\Delta BAL$. Then $\angle BAL+\angle ALB=\angle BAL+\angle AKD=90^{\circ }$, and $\Delta AMK$ has a right angle $M$, so $AL\perp DK$ (fig. 14).

Similarly $DL\perp CK$. Then lines $AL$ and $CK$, which intersect at point $P$, contain two altitudes of the $\Delta DKL$, and therefore the line $DP$ has to contain the third altitude, so $KL\perp DP$, as desired.