Ans: A
Consider $\Delta BCE$, since $BC=CE$, then $\angle CBE = \angle CEB$. Therefore, we have
Consider $\Delta BCE$, since $BC=CE$, then $\angle CBE = \angle CEB$. Therefore, we have
$\begin{array}{rcl}
\angle CBE & = & \dfrac{1}{2} ( 180^\circ – 40^\circ) \\
& = & 70^\circ
\end{array}$
Since $ABE$ is a straight line, we have
$\begin{array}{rcl}
\angle ABC & = & 180^\circ – \angle CBE \\
& = & 180^\circ – 70^\circ \\
& = & 110^\circ
\end{array}$
Since $ABCD$ is a rhombus, we have
$\begin{array}{rcl}
\angle ABC & = & \angle ADC \\
& = & 110^\circ
\end{array}$
Since $ABCD$ is a rhombus, $AD = DC$. Therefore, $\angle CAD = \angle ACD$. Hence, we have
$\begin{array}{rcl}
\angle CAD & = & \dfrac{1}{2} (180^\circ – 110^\circ) \\
& = & 35^\circ
\end{array}$
