Apply the tangent-secant theorem.

## Solution

By the property of the angle bisector of a triangle

$$
\frac{A D}{D B}=\frac{A C}{B C}=\frac{2}{3}
$$

Therefore, $A D=\frac{2}{5} A B=\frac{12}{5}$.

Let $M$ be the midpoint of $A C$. Then

$$
S_{\triangle \mathrm{ABC}}=\frac{1}{2} A C \cdot B M=\frac{1}{2} A C \sqrt{A B^{2}-A M^{2}}=\frac{1}{2} \cdot 4 \sqrt{36-4}=2 \sqrt{32}=8 \sqrt{2}
$$

By the tangent-secant theorem

$$
A M^{2}=A D \cdot A E, \text { or } 4=\frac{12}{5}\left(\frac{12}{5}-D E\right) .
$$

From this, we find that $D E=\frac{11}{15}$. Therefore,

$$
S_{\triangle \mathrm{DEC}}=\frac{D E}{A B} S_{\triangle \mathrm{ABC}}=\frac{44 \sqrt{2}}{45} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-20.jpg?height=377&width=552&top_left_y=0&top_left_x=753)

## Answer

$\frac{44 \sqrt{2}}{45}$.
[Theorem on the lengths of a tangent and a secant; the product of the entire secant and its external part. The ratio in which the bisector divides the side

In an isosceles triangle $ABC$, it is known that $AC=4$, $AB=BC=6$. The bisector of angle $C$ intersects side $AB$ at point $D$. A circle is drawn through point $D$, tangent to side $AC$ at its midpoint and intersecting segment $AD$ at point $E$. Find the area of triangle $DEC$.

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