[ Law of Sines]

On the circumcircle of triangle $ABC$, a point $M$ is taken. The line $MA$ intersects the line $BC$ at point $L$, and the line $CM$ intersects the line $AB$ at point $K$. It is known that $AL=a, BK=b, CK=c$. Find $BL$.
Apply the Law of Sines to triangles $A B L$ and $B C K$.

## Solution

Let

$$
\angle B A M=\angle B C M=\alpha, \angle A B L=\varphi
$$

By the Law of Sines in triangles $A B L$ and $B C K$, we find that

$$
\frac{B L}{\sin \alpha}=\frac{a}{\sin \varphi}, \frac{b}{\sin \alpha}=\frac{c}{\sin \left(180^{\circ}-\varphi\right)}
$$

Since $\sin \left(180^{\circ}-\varphi\right)=\sin \varphi$, by dividing the obtained equalities term by term, we find that $\frac{B L}{b}=\frac{a}{c}$, from which $B L = \frac{a b}{c}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_08342b966a7aba566077g-01.jpg?height=423&width=760&top_left_y=1338&top_left_x=655)

Answer

$\frac{a b}{c}$.