Points $B$ and $E$ lie on the circle with diameter $A C$, so $\angle B A E=\angle A B C=\angle A E C=90^{\circ}$ and $A B C E-$

is a rectangle; $A C$ is the diameter of the circle tangent to the line $C D$ at point $C$, so $\angle A C D=90^{\circ}$. Therefore, $C E$ is the altitude of the right triangle $A C D$ drawn from the vertex of the right angle, so $C D^2=D E \cdot A D$, or $108=D E(D E+8)$, from which we find that $D E=18$. Then

$$
A D=18+8=26, C E=\sqrt{A E \cdot D E}=\sqrt{8 \cdot 18}=12
$$

Therefore,

$$
S_{A B C D}=\frac{A D+B C}{2} \cdot C E=\frac{A D+A E}{2} \cdot=\frac{26+8}{2} \cdot 12=204 .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_d69082500b56170eb8ddg-07.jpg?height=660&width=1375&top_left_y=-1&top_left_x=384)

## Answer

204.00

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A circle with its center on the diagonal $A C$ of trapezoid $A B C D (B C \| A D)$ passes through vertices $A$ and $B$, is tangent to side $C D$ at point $C$, and intersects the base $A D$ at point $E$. Find the area of trapezoid $A B C D$, if $C D=6 \sqrt{13}$, $A E=8$.