Let’s draw AF perpendicular to BC. In ΔADE, as AD = AE ∠AED = ∠ADE = 60° ⇒ ΔDAE = 60° Hence, we can conclude that ΔADE is an equilateral triangle.
⇒AF=‌‌
√3
2
×4 = 2√3cm and DF = FE = 2 cm In ΔAFB, using Pythagoras’ Theorem, AB=√AF2+BF2=√(2√3)2+42=2√7cm