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

Last updated date: 25th Mar 2023

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Answer

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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.

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.

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