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

In $$\triangle{ABC}$$ and $$\triangle{ADE}$$ , we have,

$$\angle{BAD}$$ = $$\angle{EAC}$$ ...(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.