Show that the angles of an equilateral triangle are \(60^\circ\) each.

Let \(\triangle{ABC}\) be an equilateral triangle, such that

AB = BC = CA (by property)



Now, we have,
AB = AC
\(\angle{B}\) = \(\angle{C}\) ...(i)(\(\because \) angles opposite to equal sides are equal)

Similarly,
CB = CA

\(\therefore \) \(\angle{A}\) = \(\angle{B}\) ...(ii)(\(\because \) angles opposite to equal sides are equal)

\(\angle{A}\) + \(\angle{B}\) + \(\angle{C}\) = \(180^\circ\) ...(iii)(\(\because \) the sums of all angles of a triangle are \(180^\circ\))

From Equations (i),(ii) and (iii), we have,
\(\angle{A}\) + \(\angle{A}\) + \(\angle{A}\) = \(180^\circ\)
\(\Rightarrow \) 3 \(\angle{A}\) = \(180^\circ\)
\(\Rightarrow \) \(\angle{A}\) = \(60^\circ\)
\(\Rightarrow \angle{A} = \angle{B} = \angle{C} = 60^\circ\)
Hence, proved.