Q12 of 95 Page 5

ΔABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.


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


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