Section 8.4 Hyperbolic Area
¶As discussed in Section 8.3, one way to show that hyperbolic triangles with the same defect have the same area would be to show that they have the same associated Saccheri quadrilateral. It's not quite that easy, but close.
Nonetheless, consider that case first. Figure 8.4.1 shows two hyperbolic triangles with the same associated Saccheri quadrilateral. Since each triangle can be chopped up and reassembled to make the same quadrilateral, the two triangles clearly have the same area. In fact, that's the very definition of “same area.” More formally, two polygons are said to be equivalent if they can be divided into triangles such that corresponding triangles are congruent.
We can use this idea to complete the argument.
Lemma 8.4.2.
Two hyperbolic triangles with the same defect and one pair of congruent sides are equivalent.
Proof.
Since the two triangles have the same defect, their associated Saccheri quadrilaterals also have this same defect, and hence the same summit angles. Choosing to construct these quadrilaterals using the pair of congruent sides results in the two Saccheri quadrilaterals having congruent summits. As discussed in Section 8.3, the quadrilaterals are therefore congruent, so we might as well have drawn them together – as in Figure 8.4.1. But we already know that two such triangles are equivalent.
Theorem 8.4.3.
Any two hyperbolic triangles with the same defect are equivalent.
Proof.
The idea is to construct a third triangle with the same defect as the first two, but having one side congruent to each of the two triangles.
Referring yet again to Figure 8.4.1, suppose \(\triangle ABC\) is one of the two given triangles, and \(\triangle PQR\) (not shown) is the other. By bisecting line segment \(PQ\) and using a compass, it is straightforward to duplicate that line segment as \(AE\text{,}\) with \(D\) its midpoint. If we can show that \(\triangle ABE\) is the desired third triangle, we're done – using Lemma 8.4.2 twice.
All that is required to complete the proof is to show that the associated Saccheri quadrilateral of \(\triangle ABC\) is indeed the associated Saccheri quadrilateral of \(\triangle ABE\text{.}\) Yes, it looks that way, but how do we know?
What we need to show is that the intersection of line segment \(DE\) with the base (in this case, the top) of the Saccheri quadrilateral (extended, if necessary) is actually the midpoint of \(DE\text{.}\) But this follows immediately by repeated use of AAS congruence, first on the two triangles connected at point \(D\) in order to show that the perpendicular segment from \(E\) is congruent to the one from \(A\text{,}\) and then on the two triangles along line segment \(BE\text{,}\) which completes the proof.