, 2022

Let point $F$ be the midpoint of $B'C'$, let point $A'$ be the midpoint of $BC$, and let line $HF$ meet the circumcircle of $\triangle BHC'$ again at point $I$; also let line $AA'$ meet the circumcircle of $\triangle ABC$ again at point $M$. Triangle $HB'C'$ is congruent to $\triangle AB'C'$, hence similar to $\triangle ABC$. Since

$$\angle C'IF = \angle ABC = \angle A'MC,$$
$$\angle C'FI = \angle AA'B = \angle MA'C,$$
$$2C'F = C'B',$$
$$2A'C = CB,$$

we have $\triangle C'IB' \sim \triangle CMB$, and therefore $\angle FIB' = \angle A'MB = \angle ACB$. Since $\angle C'IB' = 180^\circ - \angle C'AB'$), it follows that $I$ lies on the circumcircles of $\triangle AB'C'$ and $\triangle HCB'$.

Solution 2. Denote the three interior angles of $\triangle ABC$ by $\alpha, \beta, \gamma$ respectively. It is easy to see that $\triangle ABC \sim \triangle HC'B'$. There exists a unique point $I$ inside $\triangle HC'B'$ satisfying
$$\angle HIB' = 180^\circ - \gamma, \quad \angle HIC' = 180^\circ - \beta, \quad \angle C'IB' = 180^\circ - \alpha,$$
so the three circles mentioned in the problem must all pass through point $I$.
Let line $HI$ meet $B'C'$ at point $F$. We now show that $FB' = FC''$. From $\angle HIB' + \angle HB'F = 180^\circ$, we obtain $\angle IHB' = \angle IB'F$. Similarly, $\angle IHC' = \angle IC'F$. Hence the circumcircles of $\triangle IHC'$ and $\triangle IHB'$ are tangent to $B'C'$, from which it follows that $FB'^2 = FI \cdot FH = FC'^2$.