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Ukrainian National Mathematical Olympiad

Geometry Difficulty 5.1 AIME, harder Prove it Ukraine

In isosceles triangle $ABC$ with the vertex in $B$ there are altitudes $BH$ and $CL$ . Point $D$ is such that $BDCH$ is a rectangle. Find the angle $DLH$ .
(Bogdan Rublyov)

Solution

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

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