\begin{array}{l}
\text { [The ratio in which the bisector divides the side. } \\ \text { [The ratio of the areas of triangles with a common base or common height] }]\end{array}$

In triangle $ABC$, sides $AB$ and $BC$ are equal to each other, $AC=2$, and $\angle ACB=30^{\circ}$. From vertex $A$ to the side $BC$, the bisector $AE$ and the median $AD$ are drawn. Find the area of triangle $ADE$.
The bisector of a triangle divides its side into segments proportional to the other two sides.

## Solution

Let $B H$ be the height of the given triangle, and $M$ be the projection of point $D$ onto the base. Then

$$
B H=H C \operatorname{tg} \angle B C A=\frac{\sqrt{3}}{3}, A B=B C=2 B H=\frac{2 \sqrt{3}}{3}, S_{\triangle \mathrm{ABC}}=\frac{1}{2} A C \cdot B H=\frac{\sqrt{3}}{3}
$$

By the property of the bisector of a triangle, $\frac{B E}{E C}=\frac{A B}{A C}=\frac{\sqrt{3}}{3}$, therefore

$$
\frac{B E}{B C}=\frac{\sqrt{3}}{\sqrt{3}+3}=\frac{1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{2}
$$

Since $B C=2 B D$, then $\frac{B E}{B D}=\sqrt{3}-1$. Therefore,

$$
S_{\triangle \mathrm{ABE}}=\frac{B E}{B D} \cdot S_{\Delta \mathrm{ABD}}=\frac{B E}{B D} \cdot \frac{1}{2} S_{\Delta \mathrm{ABC}}=(\sqrt{3}-1) \cdot \frac{\sqrt{3}}{6}=\frac{3-\sqrt{3}}{6}
$$

$$
S_{\triangle \mathrm{ADE}}=S_{\triangle \mathrm{ABD}}-S_{\Delta \mathrm{ABE}}=\frac{\sqrt{3}}{6}-\frac{3-\sqrt{3}}{6}=\frac{2 \sqrt{3}-3}{6}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-21.jpg?height=240&width=669&top_left_y=177&top_left_x=700)

## Answer

$\frac{2 \sqrt{3}-3}{6}$.