Question
Prove that each angle of an equilateral triangle is$ 60^\circ .$
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Answer
Given to prove each angle of an equilateral triangle is 60°. Let us consider an equilateral triangle ABC. Such that AB = BC = CA Now, AB = BC $\angle\text{A}=\angle\text{C}\ ....(\text{i})$ [Opposite angles to equal sides are equal] And $\text{BC}=\text{AC}$ $\angle\text{B}=\angle\text{A}\ ...(\text{ii)}$ From (i) and (ii), we get $\angle\text{A}=\angle\text{B}=\angle\text{C}\ ...(\text{iii})$ 
We know that Sum of angles in a triangle = 180 $\angle\text{A}+\angle\text{B}+\angle\text{C}=180$
$\angle\text{A}+\angle\text{A}+\angle\text{A}=180$
$3\angle\text{A}=60$
$\angle\text{A}=60$
$\angle\text{A}=\angle\text{B}=\angle\text{C}=60$
Hence, each angle of an equilateral triangle is $60^\circ .$
We know that Sum of angles in a triangle = 180 $\angle\text{A}+\angle\text{B}+\angle\text{C}=180$
$\angle\text{A}+\angle\text{A}+\angle\text{A}=180$
$3\angle\text{A}=60$
$\angle\text{A}=60$
$\angle\text{A}=\angle\text{B}=\angle\text{C}=60$
Hence, each angle of an equilateral triangle is $60^\circ .$
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