3. (17 points) In triangle $A B C \quad A B=4, B C=6$, angle $A B C$ is $30^{\circ}, B D-$ is the bisector of triangle $A B C$. Find the area of triangle $A B D$.
Answer: 2.4.

Solution. By the property of the angle bisector $A D: D C=4: 6=2: 3$. Therefore, $A D=\frac{2}{5}$. $A C$. The area $S_{A B D}=\frac{2}{5} S_{A B C}$, since triangles $A B C$ and $A B D$ have a common height.

![](https://cdn.mathpix.com/cropped/2024_05_06_ff9e8fc57c922dbbab4eg-2.jpg?height=294&width=354&top_left_y=133&top_left_x=934)

In the end, we get $S_{A B D}=\frac{2}{5} \cdot \frac{1}{2} \cdot A B \cdot B C \cdot \sin \angle B=\frac{1}{5} \cdot 4 \cdot 6 \cdot \frac{1}{2}=2.4$ square units.