Question

$△ABC$ is an isosceles triangle with $AC=BC.$ If $A{B}^2=2A{C}^2$ prove that $△ABC$ is a right triangle.

Answer 1

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.