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proof of AAA (hyperbolic)∗
Wkbj79†
2013-03-21 23:00:02
Following is a proof that AAA holds in hyperbolic geometry.
Proof. Suppose that we have two triangles 4ABC and 4DEF 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 `(AB) <
`(DE), where ` is used to denote length. (Note that, if `(AB) = `(DE), then
the two triangles would be congruent by ASA.) Then there are three cases:
1. `(AC) > `(DF )
2. `(AC) = `(DF )
3. `(AC) < `(DF )
Before investigating the cases, 4DEF will be placed on 4ABC 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 BC and EF
∗ hProofOfAAAhyperbolici created: h2013-03-21i by: hWkbj79i version: h39454i Privacy
setting: h1i hProofi h51M10i
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A=D
B
F
G
E
C
Note that ∠ABC and ∠CBE are supplementary. By hypothesis, ∠ABC and
∠DEF are congruent. Thus, ∠CBE and ∠DEF are supplementary. Therefore,
4BEG contains two angles which are supplementary, a contradiction.
Case 2:
A=D
B
C=F
E
Note that ∠ABC and ∠CBE are supplementary. By hypothesis, ∠ABC and
∠DEF are congruent. Thus, ∠CBE and ∠DEF are supplementary. Therefore,
4BCE 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.
A=D
B
C
E
F
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Note that ∠ABC and ∠CBE are supplementary. By hypothesis, ∠ABC and
∠DEF are congruent. Thus, ∠CBE and ∠DEF are supplementary. Similarly,
∠BCF and ∠DF E are supplementary. Thus, BCF E is a quadrilateral whose
angle sum is exactly 2π radians, a contradiction.
Since none of the three cases is possible, it follows that 4ABC and 4DEF
are congruent.
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