Given: In ∆ABC, D is the mid-point of BC and AE ⊥ BC and AC > AB
In right triangle ∆AEB, using Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AE)2 + (BE)2 = (AB)2
⇒ (AE)2 + (BD – ED)2 = (AB)2
⇒ (AE)2 + (ED)2 + (BD)2 – 2 (ED)(BD) = (AB)2
[∵ (a – b)2 = a2 + b2 – 2ab]
⇒ (AE2 + ED2) + (BD)2 – 2 (ED)(BD) = (AB)2
⇒ (AD)2 + (BD)2 – 2 (ED)(BD) = (AB)2
[∵ In right angled ∆AED, AE2 + ED2 =AD2]
Hence Proved