Let $K$ and $L$ be the midpoints of the sides $AB$ and $BC$, respectively, and $M$ and $N$ the feet of perpendiculars from $K$ and $L$, respectively, to the common tangent of the semicircles (Fig. 28). Then $\angle CAX = \angle LKM$. As $KM$ and $LN$ are radii of the semicircles, $KM = \frac{AB}{2}$ and $LN = \frac{BC}{2}$. Let $Y$ be the intersection point of line $KL$ with the common tangent of the semicircles. As triangles $KYM$ and $LYN$ are similar, $\frac{KY}{LY} = \frac{KM}{LN}$.

Fig. 28

Thus $\frac{KL+LY}{LY} = \frac{AB}{BC}$, whence $LY = \frac{KL \cdot BC}{AB-BC} = \frac{AC}{2} \cdot \frac{BC}{AC} = BC$. Therefore $\sin \angle NYL = \frac{LN}{LY} = \frac{BC}{AB} = \frac{1}{2}$, implying $\angle NYL = 30^\circ$. Consequently, $\angle CAX = \angle LKM = 90^\circ - \angle NYL = 60^\circ$.