In ∆ABC, ∠C = 90° and D is the midpoint of BC. Prove that AB2 = 4AD2 — 3AC2.

Question

Philoid · Accepted Answer

Given: C = 90° and D is the midpoint of BC .i.e. BC = 2CD

To Prove: AB² = 4AD² — 3AC²

In ∆ABC, using Pythagoras theorem we have,

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AC)² + (BC)² = (AB)²

⇒ (AC)² + (2CD)² =(AB)²

⇒ (AC)² + 4(CD)² =(AB)²

⇒ (AC)² + 4(AD² – AC²) = AB²

[∵ In right triangle ∆ACD, AD² =AC² + CD² ]

⇒ AC² +4AD² – 4AC² = AB²

⇒ 4AD² – 3AC² = AB²

or AB² = 4AD² — 3AC²

Hence Proved

In ∆ABC, ∠C = 90° and D is the midpoint of BC. Prove that AB² = 4AD² — 3AC².