In trapezoid $ABCD$, $\overline{AD}$ is perpendicular to $\overline{DC}$,
$AD = AB = 3$, and $DC = 6$. In addition, $E$ is on $\overline{DC}$, and $\overline{BE}$ is parallel to $\overline{AD}$. Find the area of $\triangle BEC$.

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 18$
Clearly, $ABED$ is a square with side-length $3.$ By segment subtraction, we have $EC = DC - DE = 6 - 3 = 3.$
The area of $\triangle BEC$ is \[\frac12\cdot EC\cdot BE = \frac12\cdot3\cdot3 = \boxed{\textbf{(B)}\ 4.5}.\]
~Aplus95 (Solution)
~MRENTHUSIASM (Revision)