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Answer :
In \(\triangle{ABC}\) and \(\triangle{ADE}\) , we have,
AB = AD ...(Given)
\(\angle{BAD}\) = \(\angle{EAC}\) ...(i)(Given)
On adding, \(\angle{DAC}\) on both sides in Eq. (i), we get,
\(\Rightarrow \) \(\angle{BAD}\) + \(\angle{DAC}\) = \(\angle{EAC}\) + \(\angle{DAC}\)
\(\Rightarrow \) \(\angle{BAC}\) = \(\angle{DAE}\)
and also, AC = AE ...(Given)
\(\therefore \) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{ADE}\) ...(By AAS congruency test)
Thus, BC = DE ...(By CPCT)
Hence, proved.