ABC is an isosceles right triangle. If $A B^{2} = 2 A C^{2}$ then right angle is at

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None of these

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Solution

The correct option is D C
Let $∠ C = 90^{\circ}$
In $Δ A C B$ ,
${(A B)}^{2} = {(A C)}^{2} + {(B C)}^{2}$
${(A B)}^{2} = {(A C)}^{2} + {(A C)}^{2}$
$A B^{2} = 2 A C^{2}$
$∴ ∠ C = 90^{\circ}$

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