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)