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

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

Philoid · Accepted Answer

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]

[∵ D is the midpoint of BC, so 2DC = BC]

Hence Proved

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