Ans: $\angle ABE=70^\circ$, $\angle BCD=40^\circ$
Since $BC//AD$, then we have
Since $BC//AD$, then we have
$\begin{array}{rcl}
\angle AEB & = & \angle CBE \\
& = & 70^\circ
\end{array}$
Since $AE = AB$, then we have
$\begin{array}{rcl}
\angle ABE & = & \angle AEB \\
& = & 70^\circ
\end{array}$
In $\Delta ABE$,
$\begin{array}{rcl}
\angle BAC & = & 180^\circ – \angle ABE – \angle AEB \\
& = & 180^\circ – 70^\circ – 70^\circ \\
& = & 40^\circ
\end{array}$
Since $ABCD$ is a parallelogram, then we have
$\begin{array}{rcl}
\angle BCD & = & \angle BAD \\
& = & 40^\circ
\end{array}$
