Balti Tee 2018 · Ülesanne 12

Despite of the logical symmetry of the picture the right angle in triangle $\triangle DEH$ is not $H$ but either $D$ or $E$. Denote by $w$ the circumcircle of the triangle $B_2HC_2$. Midperpendicular to the segment $C_2H$ is also the midperpendicular to $CC_1$ therefore it passes through the midpoint $X$ of side $BC$. By the similar reasoning the midperpendicular to $B_2H$ passes through $X$. Therefore $X$ is the center of the circle $w$. It is well known that the point which is symmetrical to the ortho-center $H$ with respect to the side $BC$ belongs to the circumcircle of the triangle $ABC$. The distance from this point to $X$ equals $XH$ due to symmetry, hence this point belongs $w$, therefore it coincides with $D$ or $E$, without loss of generality with $D$. Thus $DH \perp BC$. Finally, the centers of $w$ and circumcircle ($ABC$) belong to the mid-perpendicular of $BC$, therefore their common chord $DE$ is parallel to $BC$. Thus $\angle HDE = 90^\circ$.