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# 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.