In ΔABC, AB = AC Let ∠ABC = ∠ACB=x° and ∠BAC = y° ∴ 2x + y = 180° In ∠FDE, ∠FED=x° and ∠FDE = x - 30° (∵ DE||BC) ⇒ x+x-30 + 70 = 180° ∴ x = 70° We see that ∠FED = ∠DFE Hence, DEF is an isosceles triangle i.e. DE=DE = 4 cm.