proof of AAA (hyperbolic)
ProofOfAAAHyperbolic
2007-05-24 01:27:40
2007-09-14 11:59:30
Proof
AAA
0
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{pstricks}
\usepackage{psfrag}
\usepackage{graphicx}
\usepackage{amsthm}
\usepackage{xypic}
Following is a proof that AAA holds in hyperbolic geometry.
\begin{proof}
Suppose that we have two triangles $\triangle ABC$ and $\triangle DEF$ such that all three pairs of corresponding angles are congruent, but that the two triangles are not congruent. Without loss of generality, let us further assume that $\ell(AB)<\ell(DE)$, where $\ell$ is used to denote length. (Note that, if $\ell(AB)=\ell(DE)$, then the two triangles would be congruent by ASA.) Then there are three cases:
\begin{enumerate}
\item $\ell(AC)>\ell(DF)$
\item $\ell(AC)=\ell(DF)$
\item $\ell(AC)<\ell(DF)$
\end{enumerate}
Before investigating the cases, $\triangle DEF$ will be placed on $\triangle ABC$ so that the following are true:
\begin{itemize}
\item $A$ and $D$ correspond
\item $A$, $B$, and $E$ are collinear
\item $A$, $C$, and $F$ are collinear
\end{itemize}
Now let us investigate each case.
Case 1: Let $G$ denote the intersection of $\overline{BC}$ and $\overline{EF}$
\begin{center}
\begin{pspicture}(-2,-2)(2,2.5)
\psline(-1.5,-1.5)(0,2)
\psline(-1.5,-1.5)(1.17,-0.97)
\psline(-1.2,-0.8)(1.3,-1.3)
\psline(0,2)(1.3,-1.3)
\psdots(-1.5,-1.5)(-1.2,-0.8)(0,2)(1.17,-0.97)(1.3,-1.3)(0.407,-1.12)
\rput[b](0,2.1){$A=D$}
\rput[r](-1.25,-0.8){$B$}
\rput[r](-1.55,-1.5){$E$}
\rput[1](1.4,-0.8){$F$}
\rput[1](1.5,-1.5){$C$}
\rput[a](0.407,-1.35){$G$}
\end{pspicture}
\end{center}
Note that $\angle ABC$ and $\angle CBE$ are supplementary. By hypothesis, $\angle ABC$ and $\angle DEF$ are congruent. Thus, $\angle CBE$ and $\angle DEF$ are supplementary. Therefore, $\triangle BEG$ contains two angles which are supplementary, a contradiction.
Case 2:
\begin{center}
\begin{pspicture}(-2,-2)(2,2.5)
\psline(-1.5,-1.5)(0,2)
\psline(-1.5,-1.5)(1.35,-1.2)
\psline(-1.2,-0.8)(1.35,-1.2)
\psline(0,2)(1.35,-1.2)
\psdots(-1.5,-1.5)(-1.2,-0.8)(0,2)(1.35,-1.2)
\rput[b](0,2.1){$A=D$}
\rput[r](-1.25,-0.8){$B$}
\rput[r](-1.55,-1.5){$E$}
\rput[l](1.4,-1.2){$C=F$}
\end{pspicture}
\end{center}
Note that $\angle ABC$ and $\angle CBE$ are supplementary. By hypothesis, $\angle ABC$ and $\angle DEF$ are congruent. Thus, $\angle CBE$ and $\angle DEF$ are supplementary. Therefore, $\triangle BCE$ contains two angles which are supplementary, a contradiction.
Case 3: This is the most interesting case, as it is the one that holds in Euclidean geometry.
\begin{center}
\begin{pspicture}(-2,-2)(2,2.5)
\psline(-1.5,-1.5)(0,2)
\psline(-1.5,-1.5)(1.3,-1.3)
\psline(-1.2,-0.8)(1.17,-0.97)
\psline(0,2)(1.3,-1.3)
\psdots(-1.5,-1.5)(-1.2,-0.8)(0,2)(1.17,-0.97)(1.3,-1.3)
\rput[b](0,2.1){$A=D$}
\rput[r](-1.25,-0.8){$B$}
\rput[r](-1.55,-1.5){$E$}
\rput[1](1.4,-0.8){$C$}
\rput[1](1.5,-1.5){$F$}
\end{pspicture}
\end{center}
Note that $\angle ABC$ and $\angle CBE$ are supplementary. By hypothesis, $\angle ABC$ and $\angle DEF$ are congruent. Thus, $\angle CBE$ and $\angle DEF$ are supplementary. Similarly, $\angle BCF$ and $\angle DFE$ are supplementary. Thus, $BCFE$ is a quadrilateral whose angle sum is exactly $2\pi$ radians, a contradiction.
Since none of the three cases is possible, it follows that $\triangle ABC$ and $\triangle DEF$ are congruent.
\end{proof}