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