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 $E$. Find $AE$, given that $AK=KB=a, \angle BCK=\alpha, \angle CBE=\beta$.
The product of the entire secant and its external part for a given point and a given circle is constant.

## Solution

Let $R$ be the radius of the given circle. Since triangle $B C K$ is inscribed in the circle, then

$$
R=\frac{a}{2 \sin \alpha} \text {. }
$$

Since triangle $B C E$ is also inscribed in this circle, then

$$
C E=2 R \sin \beta=\frac{a \sin \beta}{\sin \alpha} \text {. }
$$

Since $A C \cdot A E=A B \cdot A K$, then

$$
\left(A E+\frac{a \sin \beta}{\sin \alpha}\right) A E=2 a^{2}
$$

or

$$
A E^{2}+\frac{a \sin \beta \cdot A E}{\sin \alpha}-2 a^{2}=0
$$

From which we find that

$$
A E=\frac{a}{2 \sin \alpha}\left(\sqrt{\sin ^{2} \beta+8 \sin ^{2} \alpha}-\sin \beta\right)
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-45.jpg?height=386&width=512&top_left_y=1776&top_left_x=773)

## Answer

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

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