Given: ABC is an isosceles triangle right angled at C.
Let AC = BC
In ∆ACB, using Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AC)2 + (BC)2 = (AB)2
⇒ (AC)2 + (AC)2 = (AB)2
[∵ABC is an isosceles triangle, AC = BC]
⇒ 2(AC)2 = (AB)2
Hence Proved