Ukrainian National Mathematical Olympiad

Let $K$ be the point where the common tangent touches the circle $w_2$ (Fig. 6). Consider the common tangent line to the two circles that passes through the point $Q$. Suppose it intersects the line $BK$ at a point $P$. By the properties of lines tangent to circles, $$PB=PQ=PK.$$ Therefore, $\angle BQK=90^{\circ }$. Since $AB$ is a diameter of $w_1$, we have that $\angle AQB=90^{\circ }$, and so the points $A,Q,K$ are collinear. It is given in the problem statement that $AB\perp BK$, which implies that $$A{B}^2=AQ\cdot AK\text{ and }A{C}^2=AQ\cdot AK$$ (from the properties of the right triangle $△ABK$ and the properties of secant and tangent lines to the circle $w_2$). It then follows that $AB=AC$. If $w_1$ intersects $BC$ at a point $M$, then $\angle BMA=90^{\circ }$, and so $AM$ is the altitude of the isosceles triangle $△ABC$. This proves that $BM=MC$, as required.