Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 8.6 Shortlist Prove it Baltic Way

Heights $BB_1$ and $CC_1$ of acute triangle $ABC$ intersect in point $H$ . $B_2$ and $C_2$ are points on segments $BH$ and $CH$ respectively such that $BB_2 = B_1H$ and $CC_2 = C_1H$ . Circumcircle of the triangle $B_2HC_2$ intersects circumcircle of triangle $ABC$ in points $D$ and $E$ . Prove that triangle $DEH$ is right.

Solution

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 $\angle HDE = 90^\circ$ .

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