\(\triangle{ABC}\) is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure).
Show that \(\angle{BCD}\) is a right angle.

In \(\triangle{ABC}\), we have,

AB = AC ...(i)(Given)
\(\angle{ACB}\) = \(\angle{ABC}\) ...(ii)(\(\because \) angles opposite to equal sides are equal)
Also, AB = AD ...(iii)(Given)

From (i) and (iii), AC = AD

Now, in \(\triangle{ADC}\) , we have,

AD = AC ...(proved earlier)
\(\angle{ACD}\) = \(\angle{ADC}\) ...(\(\because \) angles opposite to equal sides are equal)

Also, \(\angle{ACD}\) = \(\angle{BDC}\) ...(iv)(\(\because \angle{ADC} = \angle{BDC}\))

On adding Equations (ii) and (iv), we get,

\(\Rightarrow \angle{ACB} + \angle{ACD} = \angle{ABC} + \angle{BDC} \)
\(\Rightarrow \) \(\angle{BCD}\) = \(\angle{ABC}\) + \(\angle{BDC}\)

Adding \(\angle{BCD}\) on both sides, we have,

\( \Rightarrow \angle{BCD} + \angle{BCD} = \angle{ABC} + \angle{BDC} + \angle{BCD} \)
\(\Rightarrow \) 2 \(\angle{BCD}\) = \(180^\circ\) ...(\(\because \) sum of all angles of a triangle is \(180^\circ\))

Therefore, \(\angle{BCD}\) = \(90^\circ\)
Hence, proved.