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

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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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