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 $H' = G$. Hence $BH' \perp AB$ and $CH' \perp AC$. Also, by the choice of $A'$, $ABA'C$ is a parallelogram, meaning $A'B \parallel AC$ and $A'C \parallel AB$. Thus $A'B \perp CH'$ and $A'C \perp BH'$, which means that the reflections of $A'$ over the lines $BH'$ and $CH'$ lie on lines $A'C$ and $A'B$ respectively; denote them by $X$ and $Y$ (Fig. 12). Then $CX \parallel AB$ and $BX = BA' = AC$, analogously also $BY \parallel AC$ and $CY = CA' = AB$. 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.

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