58th Ukrainian National Mathematical Olympiad

Let us denote by $O$ the intersection of diagonals of rectangle $HBDC$ (Fig. 3). Since $△CBL$ has right angle, $BO=LO=DO$, therefore, $HO=LO=DO$. Hence, $△DHL$ is the right triangle with the hypotenuse $DH$, which yields $\angle DLH=90^{\circ }$.