Question

Prove that each angle of an equilateral triangle is ${60}^{\circ }$.

Answer 1

Let $△ABC$ be an equilateral triangle.

$\Rightarrow AB=BC=CA$



Since, AB = BC

$\Rightarrow \angle C=\angle A$

$(\because$ Angles opposite to equal sides are equal)

Similarly, $\angle A=\angle B$

$\Rightarrow \angle A=\angle B=\angle C....\left(1\right)$

We know that,

$\angle A+\angle B+\angle C={180}^{\circ }$

$\Rightarrow \angle A+\angle A+\angle A={180}^{\circ }$

$\Rightarrow 3\angle A={180}^{\circ }$

$\Rightarrow \angle A={60}^{\circ }$

$\therefore \angle A=\angle B=\angle C={60}^{\circ }$

Hence, proved