ABC is an isosceles triangle with AC = BC. If $A B^{2} = 2 A C^{2}$ , prove that $△ A B C$ is right triangle.

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Solution

Given,
$A C = B C$ and $A B^{2} = 2 A C^{2}$
Now,
$A B^{2} = 2 A C^{2}$
$A B^{2} = A C^{2} + A C^{2}$
$A B^{2} = A C^{2} + B C^{2}$
Hence, $△ A B C$ is a right triangle right-angled at C.

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ABC is an isosceles triangle with AC = BC. If ${A B}^{2} = 2 {A C}^{2}$ , prove that $Δ A B C$ is right triangle. [1 MARK]

A B C

A C = B C

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

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

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