Solution:
Let us consider that $\mathrm{ABC}$ is an equilateral triangle.
Therefore, $A B=B C=A C$
$A B=A C$
$\Rightarrow \angle C=\angle B$ (Angles opposite to equal sides of a triangle are equal)
Also,
$\mathrm{AC}=\mathrm{BC}$
$\Rightarrow \angle B=\angle A$ (Angles opposite to equal sides of a triangle are equal)
Therefore, we obtain
$\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}$
In $\triangle \mathrm{ABC}$
$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ}$
$\Rightarrow \angle \mathrm{A}+\angle \mathrm{A}+\angle \mathrm{A}=180^{\circ}$
$\Rightarrow 3 \angle \mathrm{A}=180^{\circ}$
$\Rightarrow \angle \mathrm{A}=60^{\circ}$
$\Rightarrow \angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=60^{\circ}$
Hence, in an equilateral triangle, all interior angles are of measure $60^{\circ}$.
Let us consider that $\mathrm{ABC}$ is an equilateral triangle.
Therefore, $A B=B C=A C$
$A B=A C$
$\Rightarrow \angle C=\angle B$ (Angles opposite to equal sides of a triangle are equal)
Also,
$\mathrm{AC}=\mathrm{BC}$
$\Rightarrow \angle B=\angle A$ (Angles opposite to equal sides of a triangle are equal)
Therefore, we obtain
$\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}$
In $\triangle \mathrm{ABC}$
$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ}$
$\Rightarrow \angle \mathrm{A}+\angle \mathrm{A}+\angle \mathrm{A}=180^{\circ}$
$\Rightarrow 3 \angle \mathrm{A}=180^{\circ}$
$\Rightarrow \angle \mathrm{A}=60^{\circ}$
$\Rightarrow \angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=60^{\circ}$
Hence, in an equilateral triangle, all interior angles are of measure $60^{\circ}$.
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