$\left[\begin{array}{l}{[\text { Pythagorean Theorem (direct and inverse) }} \\ {[\quad \underline{\text { Midline of a Trapezoid }}]}\end{array}\right]$

In a right trapezoid $A B C D$ ( $B C$ is parallel to $A D, A B$ is perpendicular to $A D$ ) the smaller base $A D$ is 3, and the lateral side $C D$ is 6. Point $E$, the midpoint of side $C D$, is connected to point $B$ by a straight line segment.

It is known that angle $C B E$ is $\alpha$. Find the area of trapezoid $A B C D$.
Prove that the angle at the base of the given trapezoid is $2 \boldsymbol{\alpha}$.

## Answer

Let $H$ be the midpoint of $A B$. Since $E H$ is the midline of the given trapezoid, $E H \| A D$. Therefore, $E H \perp A B$. Consequently, triangle $A B E$ is isosceles and

$$
\angle A E H=\angle B E H=\angle C B E=\alpha .
$$

From the isosceles triangle $A D E$, we find that

$$
\angle A E D=\angle E A D=\angle A E H=\alpha .
$$

Therefore,

$$
\angle D C B=\angle D E H=2 \alpha
$$

Let $D K$ be the height of the trapezoid. Then from the right triangle $D K C$, we find that

$$
D K=D C \sin \angle D C B=6 \sin 2 \alpha, C K=D C \cos \angle D C B=6 \cos 2 \alpha .
$$

Consequently,

$$
\begin{gathered}
S_{\mathrm{ABCD}}=\frac{1}{2}(A D+B C) \cdot D K=3(6+6 \cos 2 \alpha) \sin 2 \alpha= \\
\quad=18(1+\cos 2 \alpha) \sin 2 \alpha=72 \sin \alpha \cos ^{3} \alpha
\end{gathered}
$$