Show that the angles of an equilateral triangle are $60^o$ each.

(N/A) In $\Delta ABC$,we have:
$AB = BC = CA$ $[\because \Delta ABC$ is an equilateral triangle$]$
Since $AB = BC$,the angles opposite to these sides are equal,so $\angle A = \angle C$ ......... $(1)$
Similarly,since $AC = BC$,the angles opposite to these sides are equal,so $\angle A = \angle B$ ......... $(2)$
From $(1)$ and $(2)$,we have:
$\angle A = \angle B = \angle C$
Let $\angle A = \angle B = \angle C = x$
Since the sum of the angles in a triangle is $180^o$:
$\angle A + \angle B + \angle C = 180^o$
Substituting $x$ for each angle:
$x + x + x = 180^o$
$3x = 180^o$
$x = \frac{180^o}{3} = 60^o$
Therefore,$\angle A = \angle B = \angle C = 60^o$.
Thus,each angle of an equilateral triangle is $60^o$.