Given: ABC is an equilateral triangle
∴ AB = AC = BC
and AD ⊥ BC
Now, In ∆ADB, using Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AD)2 + (BD)2 = (AB)2
⇒ (AD)2 + (BD)2 = (BC)2 [∵ AB = BC]
⇒ (AD)2 + (BD)2 = (2BD)2 [ as AD⊥BC]
⇒ (AD)2 + (BD)2 = 4BD2
⇒ AD2 = 3BD2