Sine Theorem

Through the vertices $B$ and $C$ of triangle $ABC$, a circle is drawn which intersects side $AB$ at point $K$ and side $AC$ at point $L$. Find $AB$, if $AK=KB, AL=l, \angle BCK=\alpha, \angle CBL=\beta$.
If $R$ is the radius of the circle, then $R=\frac{B K}{2 \sin \angle L B C K}=\frac{C L}{2 \sin \angle L C B L}$. Further, $A K \cdot A B=A L \cdot A C$.

## Solution

Let $R$ be the radius of the circle. Denote $A K=K B=x$. Then

$$
R=\frac{B K}{2 \sin \angle B C K}=\frac{C L}{2 \sin \angle C B L}
$$

or $\frac{x}{2 \sin \alpha}=\frac{C L}{2 \sin \beta}$, from which we find that $C L=\frac{x \sin \beta}{\sin \alpha}$. By the corollary of the theorem about the tangent and the secant, $A K \cdot A B=A L \cdot A C$, or $2 x^{2}=l\left(l+\frac{x \sin \beta}{\sin \alpha}\right)$, or

$$
2 x^{2}-\frac{x l \sin \beta}{\sin \alpha}-l^{2}=0
$$

The condition of the problem is satisfied by the positive root of this quadratic equation:

$$
x=\frac{1}{4}\left(\frac{l \sin \beta}{\sin \alpha}+\sqrt{\frac{l^{2} \sin ^{2} \beta}{\sin ^{2} \alpha}+8 l^{2}}\right)=\frac{l}{4 \sin \alpha}\left(\sin \beta+\sqrt{8 \sin ^{2} \alpha+\sin ^{2} \beta}\right)
$$

Therefore, $A B=2 x=\frac{l}{2 \sin \alpha}\left(\sin \beta+\sqrt{8 \sin ^{2} \alpha+\sin ^{2} \beta}\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8027e864227276d12944g-34.jpg?height=500&width=458&top_left_y=2311&top_left_x=800)

$$
\frac{l}{2 \sin \alpha}\left(\sin \beta+\sqrt{8 \sin ^{2} \alpha+\sin ^{2} \beta}\right)
$$

Problem