3. (17 points) In triangle $ABC$, $AB=6$, $BC=9$, angle $ABC$ is $30^{\circ}$, and $BD$ is the bisector of triangle $ABC$. Find the area of triangle $ABD$.

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# Answer: 5.4.

Solution. By the property of the angle bisector $A D: D C=6: 9=2: 3$. 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-3.jpg?height=265&width=375&top_left_y=2506&top_left_x=929)

In the end, we get $S_{A B D}=\frac{2}{5} \cdot \frac{1}{2} \cdot A B \cdot B C \cdot \sin \angle B=\frac{1}{5} \cdot 6 \cdot 9 \cdot \frac{1}{2}=5.4$ sq. units.