$ABC$ is an isosceles triangle with $AC = BC$. If $AB^2 = 2 AC^2$,prove that $ABC$ is a right triangle.
(N/A) Given that,
$AB^2 = 2 AC^2$
We can rewrite this as:
$AB^2 = AC^2 + AC^2$
Since it is given that $AC = BC$,we can substitute $BC$ for one of the $AC$ terms:
$AB^2 = AC^2 + BC^2$
This equation satisfies the converse of the Pythagoras theorem,which states that if the square of one side of a triangle is equal to the sum of the squares of the other two sides,then the angle opposite the first side is a right angle.
Therefore,$\triangle ABC$ is a right-angled triangle with the right angle at $C$.
$AB^2 = 2 AC^2$
We can rewrite this as:
$AB^2 = AC^2 + AC^2$
Since it is given that $AC = BC$,we can substitute $BC$ for one of the $AC$ terms:
$AB^2 = AC^2 + BC^2$
This equation satisfies the converse of the Pythagoras theorem,which states that if the square of one side of a triangle is equal to the sum of the squares of the other two sides,then the angle opposite the first side is a right angle.
Therefore,$\triangle ABC$ is a right-angled triangle with the right angle at $C$.
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