In the fig 6.37, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.
It is given that ΔABE ≅ ΔACD.
∴ AB = AC [By cpct] ...(i)
And, AD = AE [By cpct] ...(ii)
In ΔADE and ΔABC,
AD/AB = AE/AC [Dividing equation (ii) by (i)]
∠A = ∠A [Common angle]
∴ ΔADE ~ ΔABC [By SAS similarity criterion]
