\(ABC\) is a right triangle with \(AB = AC\). Bisector of \(\angle A\) meets \(BC\) at \(D\).  Prove that  \(BC\) \(=\) \(2AD\)
 
Proof: 
 
Consider \(\triangle CAD\) and \(\triangle BAD\).
 
Thus, \(\triangle CAD \cong \triangle BAD\) [By Congruence rule]
 
So, \( = BD\) [By CPCT]
 
\(AD = BD =\) [Since midpoint of hypotenuse of a right triangle is equidistant from all the vertices]
 
\(BC = 2AD\)
 
Hence, we proved.