\(∆ABC\) is an isosceles triangle in which \(AB = AC\). Side \(BA\) is produced to \(D\) such that \(AD = AB\). Show that \(∠BCD\) is a right angle.
Proof:
In \(∆ABC\),
\(∠ACB=∠\) ------(1) [Angles opposite to equal sides are equal]
In \(∆ACD\),
\(∠ACD = ∠\) ------[2]
Adding [1] and [2] and simplifying we get,
\(∠BCD=\) \(^°\)
Therefore, \(∠BCD\) is a right angle.
Hence, we proved.