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\[ABC\] is an isosceles triangle with \[AC = BC\]. If \[A{B^2} = 2A{C^2}\], prove that \[ABC\] is a right angled triangle.

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Hint – Use Pythagoras theorem and make a diagram .

Given, \[\Delta ABC\] is an isosceles triangle .
Also given \[AC = BC....(1)\]
Also, given that
\[{(AB)^2} = 2{(AC)^2}\]
Using Pythagora's theorem, where square of hypotenuse is equal to sum of the square of other two sides of the right angled triangle.
\[\therefore \;{(AB)^2} = {(AC)^2} + {(AC)^2}.....(2)\]
Therefore,
From (1) and (2),
\[{(AB)^2} = {(AC)^2} + {(BC)^2}\]
Hence , By converse of Pythagoras theorem triangle $ABC$ is an isosceles right angled triangle.

Note - In these types of questions of isosceles triangles are those whose two sides are equal .Right angled triangle can also be isosceles . Here we have proved the questions by converse of Pythagoras theorem.