Estonian Math Competitions

Let $O$ be the circumcenter of $ABC$ and $G$ the antipode of $A$ (Fig. 11). Then $BG\perp AB$ and $CG\perp AC$. As $CH\perp AB$ and $BH\perp AC$, this means that $CG\parallel BH$ and $BG\parallel CH$. So $BHCG$ is a parallelogram. Therefore $M$, the midpoint of $BC$, is also the midpoint of $HG$. Consequently $H^{\prime }=G$. Hence $B{H}^{\prime }\perp AB$ and $C{H}^{\prime }\perp AC$. Also, by the choice of $A^{\prime }$, $AB{A}^{\prime }C$ is a parallelogram, meaning $A^{\prime }B\parallel AC$ and $A^{\prime }C\parallel AB$. Thus $A^{\prime }B\perp C{H}^{\prime }$ and $A^{\prime }C\perp B{H}^{\prime }$, which means that the reflections of $A^{\prime }$ over the lines $B{H}^{\prime }$ and $C{H}^{\prime }$ lie on lines $A^{\prime }C$ and $A^{\prime }B$ respectively; denote them by $X$ and $Y$ (Fig. 12). Then $CX\parallel AB$ and $BX=B{A}^{\prime }=AC$, analogously also $BY\parallel AC$ and $CY=C{A}^{\prime }=AB$. Therefore $ABCX$ and $ABCY$ are isosceles trapezoids. Isosceles trapezoids are cyclic quadrilaterals, thus both $X$ and $Y$ must lie on the circumcircle of $ABC$. The desired claim follows.



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Alternative solution.

Like in the previous solution, notice that $AB{A}^{\prime }C$ is a parallelogram. By symmetry with respect to $M$ notice that triangles $ABC$ and $A^{\prime }CB$ are congruent and that $H^{\prime }$ is the orthocenter of $A^{\prime }CB$. Let $X$ and $Y$ be the reflections of $A^{\prime }$ over $B{H}^{\prime }$ and $C{H}^{\prime }$ respectively (Fig. 13). Then $A^{\prime }X\perp B{H}^{\prime }$ and $A^{\prime }Y\perp C{H}^{\prime }$ and since $H^{\prime }$ is the orthocenter of $A^{\prime }CB$, we also have $A^{\prime }C\perp B{H}^{\prime }$ and $A^{\prime }B\perp C{H}^{\prime }$. So $X$ and $Y$ lie on the lines $A^{\prime }C$ and $A^{\prime }B$ respectively. The choice of $X$ and $Y$ implies that the triangles $A^{\prime }BX$ and $A^{\prime }CY$ are isosceles. Therefore $\angle A^{\prime }XB=\angle A^{\prime }YC=\angle B{A}^{\prime }C$. The desired claim follows.