\(ABC\) 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.