Question

Show that all the angles of an equilateral triangle are $60°$each.

Answer 1

Given:

An equilateral $∆\mathrm{ABC}$

To Prove:

$\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=60°$

Equilateral Triangle: The triangle in which measure of all the angles are equal and also all the sides are equal in length is known as equilateral triangle.

Based on the above definition,In$∆\mathrm{ABC}$ $\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}$$..............\left(1\right)$

By angle sum property of triangle , In$∆\mathrm{ABC}$, we have

$$\begin{gathered} \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180° \\ \angle \mathrm{A}+\angle \mathrm{A}+\angle \mathrm{A}=180° \\ 3\angle \mathrm{A}=180° \end{gathered}$$

$$\begin{gathered} \angle \mathrm{A}=\frac{{180}^0}{3} \\ ={60}^0 \end{gathered}$$

$\therefore \angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=60°$ [from equation $\left(1\right)$]

So,all the angles of an equilateral triangle are always ${60}^0$° each.

Hence Proved