[Area of a triangle (using two sides and the angle between them).]

A circle is circumscribed around triangle $ABC$. The median $AD$ is extended to intersect this circle at point $E$. It is known that $AB + AD = DE, \angle BAD = 60^{\circ}, AE = 6$. Find the area of triangle $ABC$.
On the extension of segment $E A$ beyond point $A$, lay off segment $A B_{1}$ equal to $A B$. Quadrilateral $B_{1} B E C$ is a parallelogram.

## Solution

On the extension of segment $E A$ beyond point $A$, lay off segment $A B_{1}$ equal to $A B$. Then

$$
B_{1} D = B_{1} A + A D = B A + A D = D E
$$

Therefore, quadrilateral $B_{1} B E C$ is a parallelogram. Then

$$
\angle A B C = \angle B_{1} E C = \angle B B_{1} A = 30^{\circ}, \angle A D B = 90^{\circ}
$$

Thus, $A E$ is the diameter of the circle,

$$
\angle A B E = 90^{\circ}, A C = A B = \frac{1}{2} A E = 3
$$

Therefore,

$$
S_{\triangle \mathrm{ABC}} = \frac{1}{2} A B \cdot A C \sin \angle B A C = \frac{1}{2} \cdot 3 \cdot 3 \sin 120^{\circ} = \frac{9 \sqrt{3}}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5827e7e7a03e1b3b311bg-52.jpg?height=486&width=834&top_left_y=2295&top_left_x=612)

Problem