Question

In ΔABC, D is the mid-point of BC and AE ⊥ BC . If AC > AB, show that AB² = AD² - BC .DE + 1/4 BC².

Answer 1

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