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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 triangle.

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Solution

## Proving that $△\mathrm{ABC}$ is a right angle triangle :Given:$△ABC$ is an isosceles triangle$\mathrm{AC}=\mathrm{BC}$and${\mathrm{AB}}^{2}=2{\mathrm{AC}}^{2}$To prove:$△\mathrm{ABC}$ is a right triangle.Or$A{C}^{2}+B{C}^{2}=A{B}^{2}$Proof:In $\mathrm{\Delta ACB}$,$\mathrm{AC}=\mathrm{BC}$The angles corresponding to these sides are equal so these two angles must be less than 90 degrees.From the given,$\begin{array}{rcl}{\mathrm{AB}}^{2}& =& 2{\mathrm{AC}}^{2}\\ {\mathrm{AB}}^{2}& =& {\mathrm{AC}}^{2}+{\mathrm{AC}}^{2}\end{array}$$\begin{array}{rcl}{\mathrm{AB}}^{2}& =& {\mathrm{AC}}^{2}+{\mathrm{BC}}^{2}\end{array}$ …[Since,$\mathrm{AC}=\mathrm{BC}$]By the Pythagoras theorem,$\mathrm{\Delta ABC}$ is a right-angle triangle.Hence we proved that$\mathrm{\Delta ABC}$ is a right-angle triangle.

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