6. [5 points] Two circles of the same radius 7 intersect at points $A$ and $B$. On the first circle, a point $C$ is chosen, and on the second circle, a point $D$ is chosen. It turns out that point $B$ lies on the segment $C D$, and $\angle C A D=90^{\circ}$. On the perpendicular to $C D$ passing through point $B$, a point $F$ is chosen such that $B F=B D$ (points $A$ and $F$ are on opposite sides of the line $C D$). Find the length of the segment $C F$.

![](https://cdn.mathpix.com/cropped/2024_05_06_734640e0e89abd1c02f1g-07.jpg?height=766&width=786&top_left_y=642&top_left_x=635)

Fig. 3: variant 2, problem 6
Answer: 14.

Solution. Let $R=7$ be the radii of the circles given in the condition, $\angle B A D=\alpha, \angle B C F=\beta$. Then $\angle B A C=\frac{\pi}{2}-\alpha$, and by the sine rule $B D=2 R \sin \alpha, B C=2 R \sin \left(\frac{\pi}{2}-\alpha\right)=2 R \cos \alpha$. Therefore, $C F^{2}=B C^{2}+B D^{2}=4 R^{2} \cos ^{2} \alpha+4 R^{2} \sin ^{2} \alpha=4 R^{2}$, from which $C F=2 R=14$.