Question

Prove that in a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle.

Answer 1

Given-In a $△ABC$,square of one side is equal to the sum of the square of the other two sides.$A{B}^2+B{C}^2=A{C}^2$

To prove -$\angle B={90}^{\circ }$

Construction-Draw a $△XYZ$ such that $XY=AB,YZ=BC$ and $XZ=AC$

Proof-

$\because △XYZ$ is a right angle triangle so by Pythagoras Theorem

$X{Z}^2=X{Y}^2+Y{Z}^2$

$\Rightarrow X{Z}^2=A{B}^2+B{C}^2$ $(XY=AB,YZ=BC)$

$\Rightarrow X{Z}^2=A{C}^2$ $(\because A{C}^2=A{B}^2+B{C}^2)$

$\Rightarrow XZ=AC$

$\therefore △ABC\cong △XYZ$

$\angle B={90}^{\circ }$