In figure, AC = AE, AB = AD and \(\angle{BAD}\) = \(\angle{EAC}\). Show that BC = DE.

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.