In triangle $ABC$, the sides are known: $AB=6$, $BC=4$, $AC=8$. The bisector of angle $C$ intersects side $AB$ at point $D$. A circle is drawn through points $A$, $D$, and $C$, intersecting side $BC$ at point $E$. Find the area of triangle $ADE$.

#
Triangle $A D E$ is isosceles, $\sin \angle A D E = \sin \angle A C B$.

## Solution

By the property of the angle bisector in a triangle, $\frac{A D}{D B} = \frac{A C}{B C} = \frac{2}{1}$. Therefore, $D E = A D = 4$.

Since quadrilateral $A D E C$ is cyclic, $\sin \angle A D E = \sin \angle A C B$. By the cosine rule,

$$
\cos \angle A C B = \frac{A C^2 + B C^2 - A B^2}{2 A C \cdot B C} = \frac{11}{16}
$$

Then $\sin \angle A C B = \frac{3 \sqrt{15}}{16}$. Therefore,

$$
S_{\triangle \mathrm{ADE}} = \frac{1}{2} A D \cdot D E \sin \angle A D E = \frac{1}{2} A D \cdot D E \sin \angle A C B = \frac{3 \sqrt{15}}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-05.jpg?height=460&width=458&top_left_y=2165&top_left_x=800)

## Answer

$\frac{3 \sqrt{15}}{2}$.