Question

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

Answer 1

Given : $\Delta ABC$ is an equilateral triangle

To prove : $\angle A=\angle B=\angle C={60}^{\circ }$

Proof : In $\Delta ABC,$

$AB=AC\text{(Sides of an equilateral triangle)}$

$\therefore \angle C=\angle B...\left(i\right)$

$\text{(Angles opposite to equal angles)}$

$\text{Similarly,}AB=BC$

$\therefore \angle C=\angle A...\left(ii\right)$

$From\left(i\right)and\left(ii\right)$

$\angle A=\angle B=\angle C$

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

$\left(\text{Sum of angles of a triangle}\right)$

$\therefore \angle A=\angle B=\angle C=\frac{{180}^{\circ }}{3}={60}^{\circ }$