2011-II-23

Ans: B
In $\Delta ACD$,

$\begin{array}{rcl}
\because AD & = & CD \\
\therefore \angle ACD & = & x
\end{array}$

In $\Delta BCD$,

$\begin{array}{rcl}
\because BD & = & CD \\
\therefore \angle CBD & = & y
\end{array}$

In $\Delta ABC$,

$\begin{array}{rcl}
\angle BAC + \angle ACB + \angle ABC & = & 180^\circ \\
x + (x+y) + y & = & 180^\circ \\
2x + 2y & = & 180^\circ \\
2(x+y) & = & 180^\circ \\
x + y & = & 90^\circ
\end{array}$