6. [6 points] a) Two circles of the same radius 5 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 $C F$.
b) Suppose it is additionally known that $B C=6$. Find the area of triangle $A C F$.
Answer: a) $C F=10 ;$ b) $S_{A C F}=49$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ffd4340bf000b94108d2g-05.jpg?height=777&width=791&top_left_y=357&top_left_x=632)

Fig. 3: variant 5, problem 6

Solution. a) Let $R=5$ - 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 theorem $B D=2 R \sin \alpha, B C=2 R \sin \left(\frac{\pi}{2}-\alpha\right)=2 R \cos \alpha$. 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 $C F=2 R=10$.

b) Since $\operatorname{tg} \beta=\frac{B F}{B C}=\frac{B D}{B C}=\frac{\sin \alpha}{\cos \alpha}=\operatorname{tg} \alpha$, then $\beta=\alpha$. Further, the angles $A D C$ and $A C D$ are inscribed in equal circles and subtend the same chord $A B$, so they are equal, and from the right triangle $C A D$ we find that $\angle A D C=\angle A C D=\frac{\pi}{4}$. Then $\angle A B C=$ $\pi-\frac{\pi}{4}-\left(\frac{\pi}{2}-\alpha\right)=\frac{\pi}{4}+\alpha$, so $A C=2 R \sin \left(\frac{\pi}{4}+\alpha\right)$. Therefore, $S_{A C F}=\frac{1}{2} \cdot C A \cdot C F \cdot \sin \angle A C F=$ $\frac{1}{2} \cdot 2 R \sin \left(\frac{\pi}{4}+\alpha\right) \cdot 2 R \sin \left(\frac{\pi}{4}+\beta\right)=2 R^{2} \sin ^{2}\left(\frac{\pi}{4}+\alpha\right)=R^{2}\left(1-\cos \left(\frac{\pi}{2}+2 \alpha\right)\right)=R^{2}(1+\sin 2 \alpha)=$ $R^{2}\left(1+2 \cos \alpha \sqrt{1-\cos ^{2} \alpha}\right)$, where $\cos \alpha=\frac{B C}{2 R}=\frac{3}{5}$. Thus, $S_{A C F}=49$.