In given figure ABC is an isosceles right triangle right-angled at C. Prove that ${A B}^{2} = 2 {A C}^{2}$

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Solution

Since $△ A B C$ is a right-angled triangle at C.

$∴$ ${A B}^{2} = {A C}^{2} + {B C}^{2}$

$\Rightarrow$ ${A B}^{2} = {A C}^{2} + {A C}^{2}$ [ $∵ B C = A C$ ]

$\Rightarrow$ ${A B}^{2} = {2 A C}^{2}$ $[H e n c e p r o v e d]$

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