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)


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