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Answer :
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.