In parallelogram $A B C D$, angle $B A D$ is $60^{\circ}$, and side $A B$ is 3. The bisector of angle $A$ intersects side $B C$ at point $E$. Find the area of triangle $A B E$.

#
Prove that triangle $ABE$ is isosceles.

## Solution

Since

$$
\angle BEA = \angle DAE = \angle BAE,
$$

triangle $ABE$ is isosceles,

$$
BE = AB = 3, \angle ABE = 180^\circ - \angle BAD = 120^\circ.
$$

Therefore,

$$
S_{\triangle \mathrm{ABE}} = \frac{1}{2} AB \cdot BE \sin 120^\circ = \frac{9 \sqrt{3}}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_d1936174aeda27dcc10dg-26.jpg?height=320&width=694&top_left_y=1049&top_left_x=682)

Answer

$\frac{9 \sqrt{3}}{4}$.