Ans: C
Since $XY$ and $XZ$ are the tangents to the circle $ABCD$ at $A$ and $B$ respectively, then $XA=XB$. Therefore, we have $\angle XAB = \angle XBA$. Consider $\Delta XAB$,
Since $XY$ and $XZ$ are the tangents to the circle $ABCD$ at $A$ and $B$ respectively, then $XA=XB$. Therefore, we have $\angle XAB = \angle XBA$. Consider $\Delta XAB$,
$\begin{array}{rcl}
\angle XAB & = & \dfrac{1}{2} (180^\circ – 50^\circ) \\
& = & 65^\circ
\end{array}$
Consider the straight line $XAY$,
$\begin{array}{rcl}
\angle BAD & = & 180^\circ – 65^\circ – 30^\circ \\
& = & 85^\circ
\end{array}$
Consider the cyclic quadrilateral $ABCD$,
$\begin{array}{rcl}
\angle BCD & = & 180^\circ – 85^\circ \\
& = & 95^\circ
\end{array}$
