The reflection of the orthocentre lies on the circumcircle

Idea. Show $\angle BH'C=180^\circ-\angle A$, with $A$ and $H'$ on opposite sides of $BC$: this says $ABH'C$ is a cyclic quadrilateral.

The angle $\angle BHC$. Let $E$ be the foot of the altitude from $B$ (on $AC$) and $F$ the foot of the altitude from $C$ (on $AB$). Since the triangle is acute, $H$ is inside the triangle, on segments $BE$ and $CF$. In the quadrilateral $AFHE$, the angles at $E$ and $F$ are $90^\circ$ and the angle at $A$ is $\angle A$, so
$$\angle FHE=360^\circ-90^\circ-90^\circ-\angle A=180^\circ-\angle A.$$
The angles $\angle BHC$ and $\angle FHE$ are vertical angles, hence $\angle BHC=180^\circ-\angle A$.

Reflecting. Reflection in $BC$ preserves angles, so $\angle BH'C=\angle BHC=180^\circ-\angle A$. The point $H$ is on the same side of $BC$ as $A$, so $H'$ is on the opposite side. Therefore $\angle BAC+\angle BH'C=180^\circ$ with $A,H'$ on opposite sides of $BC$: the quadrilateral $ABH'C$ is cyclic. $\blacksquare$

Remark. Consequently the circle through $B$, $H$, $C$ is the reflection of the circumcircle in $BC$, so it has the same radius $R$. Also, if $D$ is the foot of the altitude from $A$, then $D$ is the midpoint of $HH'$, so the line $AD$ meets the circumcircle again exactly at $H'$.

Lesson. When an orthocentre appears, compute $\angle BHC=180^\circ-\angle A$; reflections of $H$ in the sides land on the circumcircle.