[ The ratio in which the bisector divides the side. ] [ Angles subtending equal arcs and equal chords]

Side $A B$ of triangle $A B C$ is equal to $3, B C=2 A C, E$ is the point of intersection of the extension of the bisector $C D$ of the given triangle with the circumscribed circle around it, $D E=1$. Find $A C$.
Triangle $A D E$ is isosceles, quadrilateral $A C B E$ is an isosceles trapezoid.

## Solution

Let $A C=x$. By the property of the angle bisector of a triangle,

$$
\frac{A D}{B D}=\frac{A C}{B C}=\frac{1}{2}
$$

therefore $A D=1, B D=2$. Triangle $A D E$ is isosceles because $A D=D E=1$, so

$$
\angle A E C=\angle B A E=\angle B C E,
$$

which means $A E \| B C$. Since

$$
\angle A C E=\angle B C E=\angle B A E=\angle A E C
$$

triangle $C A E$ is also isosceles. Therefore, $A C B E$ is an isosceles trapezoid, in which

$$
B E=A C=A E=x, B C=2 x, A B=C E=3
$$

Let $A K$ be the height of trapezoid $A C B E$. Then

$$
\begin{gathered}
C K=\frac{B C-A E}{2}=\frac{2 x-x}{2}=\frac{x}{2} \\
B K=\frac{B C+A E}{2}=\frac{2 x+x}{2}=\frac{3 x}{2} \\
A K^{2}=A C^{2}-C K^{2}=x^{2}-\frac{x^{2}}{4}=\frac{3 x^{2}}{4} \\
B K^{2}+A K^{2}=A B^{2}=9, \text { or } \frac{9 x^{2}}{4}+\frac{3 x^{2}}{4}=3 x^{2}=9
\end{gathered}
$$

from which $A C=x=\sqrt{3}$.

## Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_f7f9d90261bfc416bd71g-35.jpg?height=71&width=103&top_left_y=919&top_left_x=14)