Question.
$\mathrm{ABC}$ is an isosceles triangle with $\mathrm{AC}=\mathrm{BC}$. If $\mathrm{AB}^{2}=2 \mathrm{AC}^{2}$, prove that $\mathrm{ABC}$ is a right triangle.
$\mathrm{ABC}$ is an isosceles triangle with $\mathrm{AC}=\mathrm{BC}$. If $\mathrm{AB}^{2}=2 \mathrm{AC}^{2}$, prove that $\mathrm{ABC}$ is a right triangle.
Solution:
$\mathrm{As}, \mathrm{AB}^{2}=2 \mathrm{AC}^{2}$
$\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{AC}^{2}$
$=\mathrm{AC}^{2}+\mathrm{BC}^{2} \quad[\because \mathrm{AC}=\mathrm{BC}]$
As it satisfy the pythagoran triplet
So, $\triangle \mathrm{ABC}$ is right triangle, right angled at $\angle \mathrm{C}$.
$\mathrm{As}, \mathrm{AB}^{2}=2 \mathrm{AC}^{2}$
$\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{AC}^{2}$
$=\mathrm{AC}^{2}+\mathrm{BC}^{2} \quad[\because \mathrm{AC}=\mathrm{BC}]$
As it satisfy the pythagoran triplet
So, $\triangle \mathrm{ABC}$ is right triangle, right angled at $\angle \mathrm{C}$.
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