Prove that each angle of an equilateral triangle is $60^o$.

To do:

We have to prove that each angle of an equilateral triangle is $60^o$.

Solution:

Let $\triangle ABC$ is an equilateral triangle.

In $\triangle ABC$,

$AB = AC$              (Sides of an equilateral triangle are equal)

$\angle C = \angle B$....…(i)    (Angles opposite to equal sides are equal)

Similarly,

$AB = BC$

This implies,

$\angle C = \angle A$.....…(ii)

From (i) and (ii), we get,

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

We know that,

Sum of the angles in a triangle is $180^o$.

This implies,

$\angle A + \angle B + \angle C = 180^o$

$\Rightarrow \angle A + \angle A + \angle A = 180^o$

$\Rightarrow \angle A =\frac{180^o}{3}= 60^o$

$\Rightarrow \angle A =\angle B=\angle C= 60^o$

Hence proved.