E/NE 2: Let's Not Take Sides

[stoutfellow]

Euclidean and hyperbolic geometry differ in only one respect, namely, the parallel postulate: given a line l and a point P not on l, Euclidean geometry says that there is exactly one line through P which doesn't intersect l, while hyperbolic geometry says that there are more than one. Neutral geometry is the body of results that can be proven in either kind of geometry - i.e., everything that can be proven using the axioms other than the parallel postulate (either version). Among the theorems of neutral geometry are the side-side-side, side-angle-side, and angle-side-angle theorems (conditions under which two triangles can be guaranteed to be congruent), the fact that two circles intersect in at most two points, and the fact that, given our old friends l and P, there is a line through P which is perpendicular to l. There's one more useful result, the Saccheri-Legendre theorem, which will be the subject of this post. (Saccheri was an eighteenth-century geometer who came very close to discovering non-Euclidean geometry, but couldn't quite make the intellectual leap.) This theorem says: The sum of the angles of a triangle is at most 180 degrees. (In Euclidean geometry, it's exactly 180 degrees; in hyperbolic geometry, it's less than 180, and varies depending on the triangle.) I'll put the proof under the cut.
First, we have to do some spadework. Let ABC be a triangle, and let D be a point on the other side of BC from A. Let l be the line between D and B, and consider the sum of the angles DBC and ABC. One of three things happens: I) the sum is less than 180 degrees, in which case D lies on the same side of AB as C; II) the sum is exactly 180, in which case l is AB; or III) the sum is greater than 180, in which case D is on the opposite side of AB from C and the line l enters ABC at B.

Now, let M be the midpoint of BC, and rotate ABC 180 degrees around M. B rotates to C, C rotates to B, and A rotates around to a new point D. The triangle DCB is congruent to ABC; in particular, angle DBC is equal to angle ACB, and DCB to ABC. So the sum of the angles ABC and ACB is equal to the sum of the angles ABC and DBC. Again, there are three possibilities: that sum may be less than, equal to, or greater than 180. I want to eliminate the second and third possibilities.

Suppose the sum is equal to 180. Then Case II, above, holds, and the point D lies on AB. But the sum is also equal to the sum of the angles ACB and DCB, so I can apply the same argument, swapping the roles of B and C, and so D lies on AC. So the lines AB and AC intersect in two different points, A and D. This is impossible.

Suppose the sum is greater than 180. Then Case III, above, holds, and the line BD enters ABC at B. Therefore it must exit ABC at some point E on the opposite edge, AC. But now, switching attention to C, we find that CD enters the triangle BCE at C, and it must exit through a point F on the opposite side BE. But the line BE is the line BD, and now D and F both lie on BD and CD. Again, this is impossible.

What we've just shown is this:
Lemma 1. If ABC is a triangle, the sum of the angles ABC and ACB is less than 180.

For this next part, it's probably a good idea to draw a picture. Let ABC be a triangle, and let M be the midpoint of BC. Now, the angles MAB and MAC add up to the angle BAC, which is - say - d degrees. MAB and MAC can't both be greater than d/2 degrees; one of them has to be less than or equal to d/2. Let's assume that it's MAC. (If not, just rename the vertex B "C" and the vertex C "B".) Next, as we did before, rotate ABC 180 degrees around M; A rotates to a new point D, and we're going to compare the triangles ABC and ADC. First, the triangles MAB and MDC are congruent. This is because the sides MB and MC are equal (M is the midpoint of BC); so are the sides MA and MD (we rotated A around M to get D); and finally the angles AMB and DMC are also equal. The side-angle-side theorem applies, and the two triangles are congruent. In particular, the angle MDC is equal to the angle MAB, and the angle MCD is equal to the angle MBA. Also, the angle BAC is the sum of the angles MAB and MAC, and the angle ACD is the sum of the angles MCA and MCD. (I really suggest that you draw a picture!) Now, look at the sum of the angles of ABC, and the sum of the angles of ADC. We get:
ABC+ACB+BAC = MBA+MCA+BAC = MCD+MCA+(MAB+MAC) = (MCD+MCA)+MAB+MAC = ACD+MDC+MAC = ACD+ADC+DAC.
Hence the sum of the angles of ABC equals the sum of the angles of ADC, and the angle DAC is at most half of the angle BAC.
This gives us
Lemma 2. If ABC is a triangle, there is another triangle A'B'C' with the same angle-sum, and with the angle at A' at most half of the angle at A.

Now we can prove Saccheri-Legendre. Rather than do that in full generality, let's take an example. Suppose the sum of the angles of ABC is, let's say, 190 degrees. Then the angle at A has to be greater than 10 degrees; otherwise, the sum of the angles at B and C would be greater than 180, contradicting Lemma 1. Could the angle at A be 20 degrees or less? If it were, then we can use Lemma 2 to construct A'B'C', whose angle-sum is still 190, and the angle at A' would be at most 10 degrees, which is impossible; so the angle at A has to be more than 20 degrees. Can it be 40 degrees or less? If it were, the angle at A' would be no more than 20 degrees, which we just saw was impossible. Keep going; no matter how big the angle at A is, we're going to run into a contradiction. But there's nothing special about the sum being 190 degrees; as long as the sum is greater than 180, we can pull the same kind of stunt. So the sum can't be greater than 180.


The argument under the cut is rather involved; if you have trouble following it (even after drawing some pictures), don't hesitate to say so.