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
$$AB^2 = AQ \cdot AK \text{ and } AC^2 = AQ \cdot AK$$
(from the properties of the right triangle $\triangle ABK$ 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 $\triangle ABC$. This proves that $BM = MC$, as required.