\(ABC\) and \(DBC\) are two triangles on the same base \(BC\) such that \(A\) and \(D\) lie on the opposite sides of \(BC\), \(AB = AC\) and \(DB = DC\). Show that \(AD\) is the perpendicular bisector of \(BC\).
Proof:
\(\triangle ABD \cong \triangle ACD\) [By Congruence rule]
\(\angle BAD = \angle\) [By CPCT]
\(\triangle AOB \cong \triangle AOC\) [By Congruence rule]
And, \(\angle AOB = \angle\) [By CPCT]
\(\angle AOB + \angle AOB =\) \(^{\circ}\) .
This proves that \(AD\) is the perpendicular bisector of \(BC\).
Hence, we proved.