∆ABC is an isosceles triangle, right-angled at C.Prove that AB² = 2AC².

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Solution

∆ABC is an isosceles triangle with AC = BC.
Also, ∆ABC is a right-angled triangle.
Therefore, applying Pythagoras theorem, we get:

$A B^{2} = A C^{2} + B C^{2} \Rightarrow A B^{2} = A C^{2} + A C^{2} (∵ A C = B C) \Rightarrow A B^{2} = 2 A C^{2}$

This completes the proof.

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Similar questions

$A B C$ is an isosceles triangle right angled at $C$ . Prove that $A B^{2} = 2 A C^{2}$

A B C

C .

A B^{2} = 2 A C^{2}

A B^{2} = 2 A C^{2}

{A B}^{2} = 2 {A C}^{2}

A B^{2} = 2 A C^{2}

△ A B C

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