Through point $A$, two lines are drawn: one of them touches the circle at point $B$, and the other intersects this circle at points $C$ and $D$ such that $D$ lies on the segment $A C$. Find $A B$, $C D$, and the radius of the circle, if $B C=4$, $B D=3$, and $\angle B A C=\arccos 1 / 3$.
Let $\angle B A C=\alpha, A B=x$. By the condition $\sin \alpha=\frac{2 \sqrt{2}}{3}$. From the theorem about the angle between a tangent and a chord, it follows that $\angle A B D=\angle B C D$, so triangles $A B D$ and $A C B$ are similar by two angles, with the similarity coefficient equal to ${ }^{B D} /{ }_{B C}=3 / 4$. Therefore, $A D=3 / 4 A B=3 x / 4, A C=4 / 3 A B=4 x / 3$.

By the cosine theorem, $B C^{2}=A B^{2}+A C^{2}-2 A B \cdot A C \cos \alpha$, or $16=x^{2}+{ }^{16} / 9 x^{2}-2 x^{4 x} /{ }^{2} \cdot 1 / 3={ }^{17} / 9 x^{2}$, from which $x=$ $\frac{12}{\sqrt{17}}, C D=A C-A D={ }^{7 x} / 12=\frac{7}{\sqrt{17}}$.

By the sine theorem, $\sin \angle B D C=\sin \angle A D B=\frac{A B \sin \alpha}{B D}=\frac{x \cdot \frac{2 \sqrt{2}}{3}}{3}=\frac{8 \sqrt{2}}{3 \sqrt{17}}$.

If $R$ is the radius of the circle, then $R=\frac{B C}{\sin \angle B D C}=\frac{3 \sqrt{17}}{4 \sqrt{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_95d66af6ee6f6b4b6717g-21.jpg?height=688&width=855&top_left_y=-1&top_left_x=639)

## Answer

$$
A B=\frac{12}{\sqrt{17}}, \quad C D=\frac{7}{\sqrt{17}}, \quad R=\frac{3 \sqrt{17}}{4 \sqrt{2}}
$$