2010-II-25

Ans: A
$\begin{array}{rcl}
\angle ACD & = & 180^\circ – 100^\circ \\
& = & 80^\circ
\end{array}$

In $\Delta ACD$, since $AC=AD$, then we have

$\begin{array}{rcl}
\angle ADC & = & \angle ACD \\
& = & 80^\circ
\end{array}$

Since $AB//ED$, then we have

$\begin{array}{rcl}
\angle ADE + \angle ADC + \angle ACD & = & 180^\circ \\
\angle ADE + 80^\circ + 80^\circ & = & 180^\circ \\
\angle ADE & = & 20^\circ
\end{array}$