In a right trapezoid $A B C D(A B \| C D)$ the angle at vertex $B$ measures $75^{\circ}$. Point $H$ is the foot of the perpendicular from point $A$ to the line $B C$. If $B H=D C$ and $A D+A H=8$, find the area of $A B C D$.
Produce the legs of the trapezoid until they intersect at point \( E \). The triangles \( ABH \) and \( ECD \) are congruent (ASA). The area of \( ABCD \) is equal to the area of triangle \( EAH \) of hypotenuse

\[
AE = AD + DE = AD + AH = 8
\]

Let \( M \) be the midpoint of \( AE \). Then

\[
ME = MA = MH = 4
\]

and \(\angle AMH = 30^\circ\). Now, the altitude from \( H \) to \( AM \) equals one half of \( MH \), namely 2. Finally, the area is 8.

![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-13.jpg?height=802&width=300&top_left_y=1689&top_left_x=1272)