2011-II-23
Ans: B
In $\Delta ACD$,
… Read In $\Delta ACD$,
$\begin{array}{rcl}
\because AD
$\begin{array}{rcl}
\because AD
$\begin{arra
$\because AB=BC$,
$\therefore \ang
$\begin{arr
$\begin{array}{rcl}
\dfrac{(n-2)\time
$\begin{array}{rcl}
AB^2 + BD^2
Method 1:
In $\Delta ACD$ and $\Delta BDE$,
$\angle CA