9.3. Given a trapezoid $A B C D$, where $A D \| B C, B C=A C=5$, and $A D=6$. The angle $A C B$ is twice the angle $A D B$. Find the area of the trapezoid.
# Answer: 22.

Solution: Drop a perpendicular CH from vertex C to the base AD. Denote the intersection of BD and CH as E.

Let $\angle \mathrm{ADB}=\alpha=\angle \mathrm{CBD}$ (alternate interior angles), then $\angle A C B=2 \alpha=\angle C A D$ (alternate interior angles). Triangle ACB is isosceles by condition, so $\angle A B C=\angle B A C=90 \circ-\alpha, \angle B A D=$ $\angle B A C+\angle C A D=90 \circ-\alpha+2 \alpha=90 \circ+\alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_06_bf90d7f4d154763f9290g-1.jpg?height=431&width=854&top_left_y=2020&top_left_x=1069)

Angle $\angle \mathrm{ABE}=\angle \mathrm{ABD}=\angle \mathrm{ABC}-\angle \mathrm{CBD}=90^\circ - 2 \alpha=\angle \mathrm{BCH}-\angle \mathrm{ACB}=\angle \mathrm{ACH}=\angle \mathrm{ACE}$. That is, quadrilateral $\mathrm{ABCE}$ is inscribed in a circle ( $\angle \mathrm{ABE}=\angle \mathrm{ACE}$ ), and $\mathrm{BE}$ is the diameter of this circle ( $\angle \mathrm{BCE}=\angle \mathrm{BCH}=$ $90 \circ$ ), so $\angle \mathrm{BAE}=90 \circ$, and $\angle \mathrm{EAD}=\angle \mathrm{BAD}-\angle \mathrm{BAE}=90 \circ+\alpha-90 \circ=\alpha=\angle \mathrm{ADB}=\angle \mathrm{ADE}$. Therefore, triangle AED is isosceles, EH is the altitude and median, so $\mathrm{AH}=\mathrm{HD}=\mathrm{AD}: 2=3$.

The height CH is found from the right triangle $\mathrm{ACH}: C H=\sqrt{A C^{2}-A H^{2}}=\sqrt{25-9}=\sqrt{16}=4$. The area of the trapezoid is $\frac{B C+A D}{2} \cdot C H=\frac{5+6}{2} \cdot 4=22$.

Criteria: Only the answer - 0 points. 7 points - correct answer with a correct solution. Partially correct progress in the solution - $3-5$ points.