In trapezium ABCD, $\overline{AB}\parallel \overline{CD}$. E is the midpoint of $\overline{BC}$, then $ar\left(\Delta AED\right)=$

Draw GF $\left|\right|$ AD passing through E

In AFGD,AF $\left|\right|$ DG and AD || FG

$\therefore$AFGD is a parallelogram.

$\therefore$Area of $\Delta ADE=\frac{1}{2}$area of AFGD$\to \left(1\right)$

In $\Delta BFE$ and $\Delta CGE$

$\begin{array}{l}\angle EBF=\angle ECG\\ \angle BEF=\angle CEG\\ BE=CE\left(Given\right)\\ \therefore BEF\cong \Delta \Delta CEG\\ \therefore Areaof\Delta BEF=areaof\Delta CEG\to \left(2\right)\end{array}$

From (1),area of $\Delta ADE=\frac{1}{2}\left[\text{area of AFECD\hspace{0.17em}+\hspace{0.17em}area of\hspace{0.17em}\hspace{0.17em}}\Delta CEG\right]$

$=\frac{1}{2}$[area of AFECD $+$area of $\Delta BEF$ ]from (2)

$=\frac{1}{2}$area of ABCD