In triangle $ABC$, the angle at vertex $A$ is $60^{\circ}$. A circle is drawn through points $B, C$, and point $D$, which lies on side $AB$, intersecting side $AC$ at point $E$. Find $AE$, if $AD=3, BD=1$, and $EC=4$. Find the radius of the circle.
The product of the entire secant and its external part for a given point and a given circle is constant.

Triangle $A B E$ is a right triangle.

## Solution

Let $A E=x$. The product of the entire secant and its external part for a given point and a given circle is constant, so

$$
A E \cdot A C=A D \cdot A B, \text { or } x(x+4)=12
$$

from which we find that $A E=x=2$.

In triangle $A B E$, the angle between sides $A E$ and $A B=2 A E$ is $60^{\circ}$. Therefore, this triangle is a right triangle. Then

$$
\angle B E C=\angle B E A=90^{\circ}
$$

Thus, $B C$ is the diameter of the circle. Using the cosine theorem in triangle $A B C$, we find that

$$
B C=\sqrt{A B^{2}+A C^{2}-2 A B \cdot A C \cos 60^{\circ}}=\sqrt{16+36-24}=2 \sqrt{7}
$$

Since $B C$ is the diameter of the circle, its radius is $\sqrt{7}$.

## Answer

$2 ; \sqrt{7}$