Ukrajina 2008

Let $H_2$ be a midpoint of the segment $BH$. Let $w_A$, $w_C$ be circles of radius $A_{2}A$ and $C_{2}C$ 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 $B{B}_1$. Let the second point which the circles intersect at be $B_4$. Then according to the theorem about the tangent and chord we find $\angle B_{4}AC=\angle B_{4}BA$, $\angle B_{4}CA=\angle B_{4}BC\Rightarrow \angle C{B}_{4}A=\pi -\angle B_{4}BA-\angle B_{4}BC=\pi -\angle ABC=\angle AHC$. 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 $△AHC$ into a circle circumscribed about $△A_1{B}_2{C}_1$ and respectively segment $B_{4}H$ into segment $B_3{B}_2$. In this case $B_1{B}_2$ is a diameter of the last circle which implies $B_3{B}_2\perp B{B}_1$. Therefore, $B_2$ is on the perpendicular bisector to $B{B}_4$ which has points $A_2$, $C_2$ on it as they are centers of the circles passing through $B$, $B_4$.

Fig.4