ABC is an isosceles triangle with AC = BC. If $AB^2 = 2AC^2$. Prove that ABC is a right triangle.
Given:
ABC is an isosceles triangle with AC = BC and $AB^2 = 2AC^2$.
To do:
We have to prove that ABC is a right triangle.
Solution:
In $∆ABC$,
$AB^2 = 2AC^2$.
$AB^2= AC^2 + AC^2$
$AB^2 = AC^2 + BC^2$
This implies, by converse of Pythagoras theorem,
$\angle ACB= 90^o$
Therefore,
ABC is a right triangle.
Hence proved.
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