equivalent conditions for trianglesEquivalentConditionsForTriangles2007-06-05 20:36:182007-06-06 01:24:28Theorem\usepackage{amssymb}
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\newtheorem{thm*}{Theorem}
\PMlinkescapeword{equilateral}
\PMlinkescapeword{equiangular}
The following theorem holds in Euclidean geometry, hyperbolic geometry, and spherical geometry:
\begin{thm*}
Let $\triangle ABC$ be a triangle. Then the following are equivalent:
\begin{itemize}
\item $\triangle ABC$ is \PMlinkname{equilateral}{EquilateralTriangle};
\item $\triangle ABC$ is \PMlinkname{equiangular}{EquiangularTriangle};
\item $\triangle ABC$ is \PMlinkname{regular}{RegularTriangle}.
\end{itemize}
\end{thm*}
Note that this statement does not generalize to any polygon with more than three sides in any of the indicated geometries.
\begin{proof}
It suffices to show that $\triangle ABC$ is equilateral if and only if it is equiangular.
Sufficiency: Assume that $\triangle ABC$ is equilateral.
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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.
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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.
\end{proof}