∆ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ∆ABC is a right triangle

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Solution

Given:
AC = BC
$Also, A B^{2} = 2 A C^{2} = A C^{2} + A C^{2} A B^{2} = A C^{2} + B C^{2} (∵ A C = B C)$

It is evident that the sum of squares of the two sides of triangle ABC is equal to square of the third side.

Therefore, by applying Pythagoras theorem, we conclude that ∆ABC is right-angled at C.

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Similar questions

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

△ A B C

A B C

A C = B C

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

A B C

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

Δ

=

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

Δ

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