Question 184
Let $ABCD$ be a convex quadrilateral. Suppose that the area of $ABCD$ is equal to
$$
\frac{AB+CD}{2} \cdot \frac{AD+BC}{2}
$$
Prove that $ABCD$ is a rectangle.
$$
\frac{AB+CD}{2} \cdot \frac{AD+BC}{2}
$$
Prove that $ABCD$ is a rectangle.