Proof. It suffices to show that
$\triangle ABC$ is equilateral if and only if it is equiangular.
Sufficiency: Assume that $\triangle ABC$ is equilateral.
Since $\overline{AB} \cong \overline{AC} \cong \overline{BC}$ , SSS yields that $\triangle ABC \cong \triangle BCA$ . By CPCTC, $\angle A \cong \angle B \cong \angle C$ . Hence, $\triangle ABC$ is equiangular.
Necessity: Assume that $\triangle ABC$ is equiangular.
By the theorem on determining from angles that a triangle is isosceles, we conclude that $\triangle ABC$ is isosceles with legs $\overline{AB} \cong \overline{AC}$ and that $\triangle BCA$ is isosceles with legs $\overline{AC} \cong \overline{BC}$ . Thus, $\overline{AB} \cong \overline{AC} \cong \overline{BC}$ . Hence, $\triangle ABC$ is equilateral. 
 \psline(1.15,2.2)(1.35,2.1) \psline(3.65,2.1)(3.85,2.2) \end{pspicture}](http://images.planetmath.org:8080/cache/objects/9536/js/img2.png "\begin{pspicture}(-0.2,-0.2)(5.2,5.2) \pspolygon(0,0)(5,0)(2.5,4.33) \rput[b](2.... ...2.5,0.2) \psline(1.15,2.2)(1.35,2.1) \psline(3.65,2.1)(3.85,2.2) \end{pspicture}")
{0.5}{120}{180} \psarc(2.5,4.33){0.5}{240}{300} \end{pspicture}](http://images.planetmath.org:8080/cache/objects/9536/js/img3.png "\begin{pspicture}(-0.2,-0.2)(5.2,5.2) \pspolygon(0,0)(5,0)(2.5,4.33) \rput[b](2.... ...}{60} \psarc(5,0){0.5}{120}{180} \psarc(2.5,4.33){0.5}{240}{300} \end{pspicture}")