Proof I Let $B_0$, $C_0$ be the midpoints of $CA$, $AB$ respectively. Denote $A'$ as the other intersection of the circle centered at $B_0$ which passes through $H$ and the circle centered at $C_0$ which passes through $H$. We know that $A'H \perp C_0B_0$. Since $B_0$, $C_0$ are the midpoints of $CA$, $AB$ respectively, $B_0C_0 \parallel BC$. Therefore $A'H \perp BC$. This yields that $A'$ lies on the segment $AH$.

By the Secant-Secant theorem, it follows that
$$AC_1 \cdot AC_2 = AA' \cdot AH = AB_1 \cdot AB_2,$$
So $B_1$, $B_2$, $C_1$, $C_2$ are concyclic.

Let the intersection of the perpendicular bisectors of $B_1B_2$, $C_1C_2$ be $O$. Then $O$ is the circumcenter of quadrilateral $B_1B_2C_1C_2$, as well as the circumcenter of $\triangle ABC$. So
$$OB_1 = OB_2 = OC_1 = OC_2.$$
Similarly,
$$OA_1 = OA_2 = OB_1 = OB_2.$$
Therefore, six points $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$ are all on the same circle, whose center is $O$, and radius $OA_1$.

Proof II Let $O$ be the circumcenter of triangle $ABC$, and $D$, $E$, $F$ the midpoints of $BC$, $CA$, $AB$ respectively. $BH$ intersects $DF$ at point $P$, then $BH \perp DF$. By Pythagoras' theorem, we have
$$BF^2 - FH^2 = BP^2 - PH^2 = BD^2 - DH^2, \qquad \textcircled{1}$$
$$BO^2 - A_1O^2 = BD^2 - A_1D^2 = BD^2 - DH^2. \qquad \textcircled{2}$$
Similarly,
$$BO^2 - C_2O^2 = BF^2 - FH^2, \qquad \textcircled{3}$$
By ①, ②, ③, $A_1O = C_2O$. It is obvious that $A_1O = A_2O$, $C_1O = C_2O$, thus
$$A_1O = A_2O = C_1O = C_2O.$$
Similarly,
$$A_1O = A_2O = B_1O = B_2O.$$
Therefore, six points $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$ are all on the same circle, whose center is $O$.