[ Inscribed and Circumscribed Circles] Law of Sines $\quad]$

The radius of the circle circumscribed around triangle $K L M$ is $R$. Through vertex $L$, a line perpendicular to side $K M$ is drawn. This line is intersected at points $A$ and $B$ by the perpendicular bisectors of sides $K L$ and $L M$. It is known that $A L=a$. Find $B L$.
Consider the case when triangle KLM is acute-angled. Let $O$ be the center of its circumscribed circle. Denote $\angle L K M=\alpha$. Then $\angle M O L=2 \alpha$ as a central angle corresponding to the inscribed angle $L K M, \mathrm{a} \angle$ $L O B=\frac{1}{2} \angle M O L=\alpha$.

On the other hand

$$
\angle O A L=90^{\circ}-\angle A L K=90^{\circ}-\left(90^{\circ}-\alpha\right)=\alpha
$$

therefore, triangles $O A L$ and $B O L$ are similar by two angles, which means $\frac{B L}{O L}=\frac{O L}{A L}$, hence, $B L=\frac{O L^{2}}{A L}=\frac{R^{2}}{a}$.

Similarly for an obtuse-angled triangle.

Answer

## $\frac{R^{2}}{a}$.