Why would the reflections of the orthocentre lie on the circumcircle?

Let ABC be a triangle which it not right-angled. Define a sequence of triangles $A_iB_iC_i$,with $i \geq 0$, as follows: $A_0B_0C_0$ is the triangle $ABC$; and, for $i \geq 0$, $A_{i+1}$, $B_{i+1}$, and $C_{i+1}$ are the reflections of the orthocentre of triangle $A_iB_iC_i$ in the sides $B_iC_i$, $C_iA_i$, and $A_iB_i$ , respectively. Assume that $\angle A_m = \angle A_n$ for some distinct natural numbers $m, n$. Prove that $\angle A = 60$.

Why would the reflections of the orthocentre lie on the circumcircle?

this is the 5th question of crmo

Let $Q$ be the orthocenter of $\triangle ABC$, and let $Q^\prime$ be the reflection of $Q$ in side $\overline{BC}$.

Note that $\angle QCA$ and $\angle A$ are complements (as acute angles of a right triangle); likewise, $\angle QAC$ and $\angle C$ are complements.

Then, $$\angle Q^\prime = \angle Q^\prime QC = \angle QAC + \angle QCA = \left(\; 90^\circ - C \;\right) + \left(\; 90^\circ - A \;\right) = 180^\circ - A - C = B$$

Since points $B$ and $Q^\prime$ subtend the same angle with segment $\overline{AC}$, they lie on the same circle through $A$ and $C$. (This is an aspect of the Inscribed Angle Theorem.) Thus, the reflection of the orthocenter in a side of a triangle lies on the circumcircle of that triangle.