8,9 The diagonal $AC$ of the isosceles trapezoid $ABCD$ is equal to $a$ and forms angles $\alpha$ and $\beta$ respectively with the larger base $AD$ and the lateral side $AB$. Find the bases of the trapezoid.
Let $x$ and $y$ be the bases of the trapezoid and form a system of equations with respect to $x$ and $y$.

## Solution

Let $B C=x, A D=y$. Suppose $C H$ is the height of the trapezoid. Then

\[
\begin{gathered}
A H=A C \cos \angle C A H=a \cos \alpha, C H=A C \sin \angle C A H=a \sin \alpha \\
D H=C H \operatorname{ctg} \angle C D H=a \sin \alpha \operatorname{ctg}(\alpha+\beta)
\end{gathered}
\]

Since

\[
D H=\frac{A D-B C}{2}=\frac{y-x}{2}, A H=\frac{A D+B C}{2}=\frac{y+x}{2},
\]

we have the system

\[
\left\{\begin{array}{l}
\frac{y-x}{2}=a \sin \alpha \operatorname{ctg}(\alpha+\beta) \\
\frac{y+x}{2}=a \cos \alpha
\end{array}\right.
\]

from which we find that

\[
x=a(\cos \alpha-\sin \alpha \cdot \operatorname{ctg}(\alpha+\beta)), y=a(\cos \alpha+\sin \alpha \cdot \operatorname{ctg}(\alpha+\beta))
\]

## Answer

$a(\cos \alpha \pm \sin \alpha \cdot \operatorname{ctg}(\alpha+\beta))$