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[parent] proof of AAA (hyperbolic) (Proof)

Following is a proof that AAA holds in hyperbolic geometry.

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:
  1. $ \ell(AC)>\ell(DF)$
  2. $ \ell(AC)=\ell(DF)$
  3. $ \ell(AC)<\ell(DF)$

Before investigating the cases, $ \triangle DEF$ will be placed on $ \triangle ABC$ so that the following are true:

  • $ A$ and $ D$ correspond
  • $ A$, $ B$, and $ E$ are collinear
  • $ A$, $ C$, and $ F$ are collinear

Now let us investigate each case.

Case 1: Let $ G$ denote the intersection of $ \overline{BC}$ and $ \overline{EF}$


\begin{pspicture}(-2,-2)(2,2.5) \psline(-1.5,-1.5)(0,2) \psline(-1.5,-1.5)(1.17,... ....4,-0.8){$F$} \rput[1](1.5,-1.5){$C$} \rput[a](0.407,-1.35){$G$} \end{pspicture}

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{pspicture}(-2,-2)(2,2.5) \psline(-1.5,-1.5)(0,2) \psline(-1.5,-1.5)(1.35,... ...5,-0.8){$B$} \rput[r](-1.55,-1.5){$E$} \rput[l](1.4,-1.2){$C=F$} \end{pspicture}

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{pspicture}(-2,-2)(2,2.5) \psline(-1.5,-1.5)(0,2) \psline(-1.5,-1.5)(1.3,-... ...-1.55,-1.5){$E$} \rput[1](1.4,-0.8){$C$} \rput[1](1.5,-1.5){$F$} \end{pspicture}

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.

$ \qedsymbol$



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Cross-references: radians, angle sum, quadrilateral, Euclidean geometry, contradiction, contains, hypothesis, supplementary, intersection, collinear, ASA, length, without loss of generality, congruent, angles, triangles, hyperbolic geometry, AAA

This is version 3 of proof of AAA (hyperbolic), born on 2007-05-24, modified 2007-09-14.
Object id is 9454, canonical name is ProofOfAAAHyperbolic.
Accessed 756 times total.

Classification:
AMS MSC51M10 (Geometry :: Real and complex geometry :: Hyperbolic and elliptic geometries and generalizations)

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