Problem 1657

Let $H_2$ be a midpoint of the segment $BH$. Let $w_A$, $w_C$ be circles of radius $A_2A$ and $C_2C$ with centers at points $A_2$, $C_2$ respectively. Then they are tangent to line $AC$ at endpoints of the segment $AC$ and pass through point $B$, since points $A_2$, $C_2$ are on the respective perpendicular bisectors. Furthermore, power of point $B_1$ about these circles is equal and therefore the radical axis of the circles is the line $BB_1$. Let the second point which the circles intersect at be $B_4$. Then according to the theorem about the tangent and chord we find $\angle B_4AC = \angle B_4BA$, $\angle B_4CA = \angle B_4BC \Rightarrow \angle CB_4A = \pi - \angle B_4BA - \angle B_4BC = \pi - \angle ABC = \angle AHC$. Then points $A$, $B_4$, $C$, $H$ are on one circle. Dilation with center $B$ and ratio $\frac{1}{2}$ (fig.4) converts the circle circumscribed about $\triangle AHC$ into a circle circumscribed about $\triangle A_1B_2C_1$ and respectively segment $B_4H$ into segment $B_3B_2$. In this case $B_1B_2$ is a diameter of the last circle which implies $B_3B_2 \perp BB_1$. Therefore, $B_2$ is on the perpendicular bisector to $BB_4$ which has points $A_2$, $C_2$ on it as they are centers of the circles passing through $B$, $B_4$.

Fig.4