Around triangle $A B C$, in which $B C=a, \angle B=\alpha, \angle C=\beta$, a circle is described. The bisector of angle $A$ intersects this circle at point $K$. Find $A K$.
A side of a triangle is equal to the diameter of the circumscribed circle multiplied by the sine of the opposite angle.

## Solution

Let $R$ be the radius of the circle circumscribed around triangle $ABC$. Then

$$
\begin{gathered}
\angle BAC = 180^{\circ} - (\beta + \alpha), \quad 2R = \frac{BC}{\sin (\beta + \alpha)} = \frac{a}{\sin (\beta + \alpha)} \\
\angle ABK = \angle ABC + \angle CBK = \alpha + \frac{1}{2} \left(180^{\circ} - (\alpha + \beta)\right) = 90^{\circ} + \frac{\alpha - \beta}{2}
\end{gathered}
$$

Therefore,

$$
\begin{gathered}
AK = 2R \sin \angle ABK = 2R \sin \left(90^{\circ} + \frac{\alpha - \beta}{2}\right) = \\
= 2R \cos \frac{\beta - \alpha}{2} = \frac{a \cos \frac{\beta - \alpha}{2}}{\sin (\beta + \alpha)}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-40.jpg?height=309&width=378&top_left_y=1046&top_left_x=840)

Answer

$\frac{a \cos \frac{\beta - \alpha}{2}}{\sin (\beta + \alpha)}$.