6. [6 points] a) Two circles of the same radius 13 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 the same side of the line $C D$). Find the length of the segment $C F$. b) Suppose it is additionally known that $B C=10$. Find the area of triangle $A C F$.
Answer: $C F=26, S_{\triangle A C F}=119$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c4692c563a78e68b8782g-08.jpg?height=477&width=782&top_left_y=904&top_left_x=637)

Fig. 3: variant 4, problem 6

Solution. a) Let $R=13$ 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 $C F=2 R=26$.

b) Since $\operatorname{tg} \beta=\frac{B F}{B C}=\frac{B D}{B C}=\frac{\sin \alpha}{\cos \alpha}=\operatorname{tg} \alpha$, then $\beta=\alpha$. Note that $\cos \beta=\frac{B C}{F C}=\frac{5}{13}$, so $\beta>\frac{\pi}{4}$. 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, $S_{A C F}=\frac{1}{2} \cdot C A \cdot C F \cdot \sin \angle A C F=\frac{1}{2} \cdot 2 R \sin \left(\frac{\pi}{4}+\alpha\right) \cdot 2 R \sin \left(\beta-\frac{\pi}{4}\right)=2 R^{2} \sin \left(\frac{\pi}{4}+\alpha\right) \sin \left(\alpha-\frac{\pi}{4}\right)=$ $-2 R^{2} \sin \left(\frac{\pi}{4}+\alpha\right) \cos \left(\alpha+\frac{\pi}{4}\right)=-R^{2} \sin \left(\frac{\pi}{2}+2 \alpha\right)=-R^{2} \cos 2 \alpha=R^{2}\left(1-2 \cos ^{2} \alpha\right)=119$.