In Fig. 6.37, if Δ ABE ≅Δ ACD, show that
Δ ADE ~ Δ ABC.

It is given in the question that ΔABE ≅ ΔACD

∴ AB = AC (By CPCT) (i)

And,

AD = AE (By CPCT) (ii)

In ΔADE and ΔABC,

Dividing equation (ii) by (i)

∠A = ∠A (Common)

Hence,

ΔADEΔABC (By SAS similarity)