equivalent conditions for triangles
The following theorem holds in Euclidean geometry^{}, hyperbolic geometry, and spherical geometry:
Theorem 1.
Let $\mathrm{\u25b3}\mathit{}A\mathit{}B\mathit{}C$ be a triangle^{}. Then the following are equivalent^{}:

•
$\mathrm{\u25b3}ABC$ is equilateral (http://planetmath.org/EquilateralTriangle);

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$\mathrm{\u25b3}ABC$ is equiangular (http://planetmath.org/EquiangularTriangle);

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$\mathrm{\u25b3}ABC$ is regular^{} (http://planetmath.org/RegularTriangle).
Note that this statement does not generalize to any polygon^{} with more than three sides in any of the indicated geometries.
Proof.
It suffices to show that $\mathrm{\u25b3}ABC$ is equilateral if and only if it is equiangular.
Sufficiency: Assume that $\mathrm{\u25b3}ABC$ is equilateral.
Since $\overline{AB}\cong \overline{AC}\cong \overline{BC}$, SSS yields that $\mathrm{\u25b3}ABC\cong \mathrm{\u25b3}BCA$. By CPCTC, $\mathrm{\angle}A\cong \mathrm{\angle}B\cong \mathrm{\angle}C$. Hence, $\mathrm{\u25b3}ABC$ is equiangular.
Necessity: Assume that $\mathrm{\u25b3}ABC$ is equiangular.
By the theorem on determining from angles that a triangle is isosceles, we conclude that $\mathrm{\u25b3}ABC$ is isosceles with legs $\overline{AB}\cong \overline{AC}$ and that $\mathrm{\u25b3}BCA$ is isosceles with legs $\overline{AC}\cong \overline{BC}$. Thus, $\overline{AB}\cong \overline{AC}\cong \overline{BC}$. Hence, $\mathrm{\u25b3}ABC$ is equilateral. ∎
Title  equivalent conditions for triangles 

Canonical name  EquivalentConditionsForTriangles 
Date of creation  20130322 17:12:46 
Last modified on  20130322 17:12:46 
Owner  Wkbj79 (1863) 
Last modified by  Wkbj79 (1863) 
Numerical id  10 
Author  Wkbj79 (1863) 
Entry type  Theorem 
Classification  msc 5100 
Related topic  Triangle 
Related topic  IsoscelesTriangle 
Related topic  EquilateralTriangle 
Related topic  EquiangularTriangle 
Related topic  RegularTriangle 