Today is a spherical geometry day. There are six theorems to cover today -- three "scholia" and three main theorems, including the big deal theorem that gives the ultimate link between angle measure and area.
Let's begin with Proposition 496, which is considered by Legendre to be a mere "scholium":
496. Scholium. The spherical wedge comprehended by the planes AMB, ANB is to the entire sphere as the angle A is to four right angles. For, the lunary surfaces being equal, the spherical wedges will also be equal; therefore two spherical wedges are to each other as the angles formed by the planes which comprehend them.
The final Legendre theorem we discussed in our last spherical geometry post compared the area of a lune to the surface area of the entire sphere. This scholium follows up with a comparison of the volume of a spherical wedge to that of the entire sphere. A spherical "wedge" is exactly what you think it is -- it's just like the wedge of an orange. A wedge is closely related to a lune -- if we start on the surface of a lune, any point we can reach by digging directly down from the lune towards the center of the sphere is in the wedge.
Legendre tells us that the volume of a wedge is exactly what you expect it to be -- take the volume of the sphere and multiply it by the measure of its angle divided by "four right angles" -- that is, 360 degrees or 2pi (sorry, make that tau, or 4lambda) radians.
We proceed with Proposition 497, our first main theorem:
497. Two symmetrical spherical triangles are equal in surface.
Now Legendre uses the word "symmetrical" in different ways throughout his text. Sometimes he uses it to mean merely "congruent," but at other times he uses it to denote the existence of a certain reflection mapping one to the other. For example, earlier in the text (before the unit on spherical geometry), Legendre calls two pyramids S-ABC and T-ABC "symmetrical." In this case the two pyramids have the same base ABC, and a reflection in the plane ABC maps S to T. In the first paragraph of his proof, Legendre explains exactly what he means by "symmetrical" here:
Demonstration. Let ABC, DEF, be two symmetrical triangles, that is, two triangles which have their sides equal, namely, AB = DE, AC = DF, CB = EF, and which at the same time do not admit of being applied one to the other; we say that the surface ABC is equal to the surface DEF.
Clearly Legendre is using "symmetrical" here to mean "congruent." After all, he lists three pairs of congruent sides, and he's already proved the spherical equivalent of SSS, so we know that the two triangles are congruent. Since he writes that the two triangles "do not admit of being applied one to the other," we assume that there is not necessarily any reflection (or any other isometry) mapping ABC to DEF (though there could be).
Legendre says that he is to prove that ABC and DEF are equal in "surface" -- that is to say, that they are equal in area. This reminds us of the Area Postulate found in Lesson 8-3 of the U of Chicago text:
c. Congruence Property: Congruent figures have the same area.
Given this postulate, it appears to be obvious that ABC and DEF have the same area -- we just said that the triangles are congruent, and the postulate tells us that congruent figures have the same area. so therefore the triangles have the same area, right? So what is there for Legendre to prove?
Here's the problem -- the postulate tells us that "congruent" figures have the same area, but what does it mean for two figures to be "congruent"? The U of Chicago text uses the Common Core definition of congruence -- that is, the existence of an isometry mapping one to the other. But Legendre expressly mentioned that we don't have such an isometry available! He tells us that the two triangles are "congruent" by SSS -- but his proof of SSS only tells us that two triangles with equal sides have equal angles, not that any isometry exists between them! In other words, the two triangles are congruent by the old pre-Core definition, but not necessarily by the Common Core definition.
Maybe we should be consistent about our own use of terminology. Let's call two triangles ABC and DEF "symmetrical," just as Legendre does, and reserve the word "congruent" only when there's a proved isometry between the two triangles. As it turns out, in this proof Legendre will ultimately divide the triangles ABC and DEF into smaller triangles, and then he does prove the existence of isometries between pairs of these smaller triangles. This isn't enough to prove that there's an isometry between the entire triangles ABC and DEF, but it is sufficient to prove (via the Additive Property of Area) that the triangles have the same area.
Okay, so now let's proceed with Legendre's proof.
"Let P be the pole of the small circle which passes through the three points A, B, C; from this point draw equal arcs PA, PB, PC (464); at the point F make the angle DFQ = ACP, the arc FQ = CP, and join DQF = APC."
Let's see what's going on so far here in Legendre. We know that a circle passes through A, B, and C (since a circle passes through any three points) -- the only fear is that the circle might be a great circle instead of a small circle. Legendre tells us in a footnote that if ABC can't be a great circle -- basically because otherwise ABC couldn't be a triangle.
Now any circle, small or great, has a pole (two poles, in fact). The pole of any circle is essentially the center of that circle. On the earth, the poles of the Equator, or any parallel of latitude, are the actual North and South Poles. The poles of a circle are equidistant from all points on the circle. (Note that Proposition 464 mentioned here details how to find the pole of a circle -- the poles are the endpoints perpendicular to the plane containing the circle.) The rest of this part of the proof is the construction of a certain triangle on side DF, the new triangle DFQ.
"The sides DF, FQ, are equal to the sides AC, CP; the angle DFQ = ACP; consequently the two triangles are equal in all their parts (480); therefore the side DQ = AP, and the angle DQF = APC.
"In the proposed triangles DFE, ABC, the angles DFE, ACB, opposite to the equal sides DE, AB, being equal (481), if we subtract from them the angles DFQ, ACP equal, by construction, there will remain the angle QFE equal to PCB. Moreover, the sides QF, FE are equal to the sides PC, CB; consequently the two triangles FQE, CPB, are equal in all their parts; therefore the side QE = PB, and the angle FQE = CPB.
"If we observe, now, that the triangles DFQ, ACP, which have the sides equal each to each, are at the same time isosceles, we shall perceive that they may be applied one to the other; for having placed PA upon its equal QF, the side PC will fall upon its equal QD, and thus the two triangles will coincide; consequently they are equal, and the surface DQF = APC. For a similar reason the surface FQE = CPB, and the surface DQE = APB; we have, accordingly,
DQF + FQE - DQE = APC + CPB - APB, or DEF = ABC; therefore the two symmetrical triangles ABC, DEF, are equal in surface. QED"
Let's move on to Proposition 498, which is considered by Legendre to be a mere "scholium":
498. Scholium. The poles P and Q may be situated within the triangles ABC, DEF; then it would be necessary to add the three triangles DQF, FQE, DQE, in order to obtain the triangle DEF, and also the three triangles APC, CPB, APB, in order to obtain the triangle ABC. In other respects the demonstration would always be the same and the conclusion the same.
Notice that Legendre's P is the pole (center) of the circle containing the points A, B, and C -- that is, P is the circumcenter of triangle ABC. In Euclidean geometry the circumcenter of ABC lies inside the triangle if it's acute, and outside the triangle if it's obtuse. If ABC is a right triangle, then the circumcenter lies on the triangle -- indeed, it's the midpoint of the hypotenuse.
This case is quite suited for a two-column proof. Here is the beginning of such a proof (which I admit will be long):
Given: AB = DE, AC = DF, CB = EF, P pole of ABC lies inside triangle ABC
Prove: Area(ABC) = Area(DEF)
1. AB = DE, AC = DF, CB = EF, P pole 1. Given
2. PA = PB = PC 2. Definition of pole of a circle
3. Q s.t. DFQ = ACP, FQ = CP 3. Ruler/Protractor Postulates
4. Triangle DFQ = ACP 4. SAS
5. DQ = AP, DQF = APC 5. CPCTC
and so on.
We proceed with Proposition 499, our second main theorem:
499. If two great circles AOB, COD, cut each other in any manner in the surface of a hemisphere AOCBD, the sum of the opposite triangles AOC, BOD, will equal to the lunary surface of which the angle is BOD.
OK, so we're getting closer to the connection between angle measure and area. Notice that Legendre clearly intends A and B to be antipodal points since he calls AOB a "semicircumference" later on, and likewise C and D are antipodal points.
Let's look at Legendre's proof now:
Demonstration. By producing the arcs OB, OD, into the surface of the hemisphere until they meet in N [that is, O and N are antipodal points -- dw], OBN will be a semicircumference as well as AOB; taking from each OB, we shall have BN = AO. For a similar reason DN = CO, and BD = AC; consequently the two triangles AOC, BDN [I prefer writing the second triangle as BND -- dw], have the three sides of the one respectively equal to the three sides of the other; moreover, their position is such that they are symmetrical; therefore they are equal in surface (496) [Legendre clearly means 497 here -- dw] and the sum of the triangles AOC, BOD is equivalent to the lunary surface OBNDO, of which the angle is BOD. QED
As usual, this is easier to visualize if we let N and O denote the North and South Poles. Notice that if A and C are in the Southern Hemisphere, their antipodes B and D are in the Northern Hemisphere (so the triangle BOD is a rather large triangle). The Southern Hemisphere triangle AOC and the Northern Hemisphere triangle BND are congr -- er, symmet -- er, congruent. (Notice that the antipodal map -- that is, the function mapping every point to its antipodes -- is in fact an isometry!) And so we fit BND and BOD together to form a lune.
Let's move on to Proposition 500, which is considered by Legendre to be a mere "scholium":
500. Scholium. It is evident, also, that the two spherical pyramids, which have for their bases the triangles AOC, BOD, taken together, are the spherical wedge of which the angle is BOD.
Recall that a spherical pyramid is formed by taking a triangle and digging towards the center of the sphere, just as a spherical wedge is formed by taking a lune and digging down. Therefore Proposition 500 follows from 499 exactly as Proposition 496 follows from 495.
We proceed with Proposition 501, our third main theorem:
501. The surface of a spherical triangle has for its measure the excess of the sum of the three angles over two right angles.
And this is the amazing link between angle measure and area that we've been preparing for! We already know from earlier that the sum of the angles of a triangle in spherical geometry is always greater than 180 degrees. This theorem tells us that the triangle sum is greater for larger triangles than it is for smaller triangles -- in fact, the area of the triangle is exactly proportional to the number of degrees past 180 that the sum of the angles is. And if we choose the correct units -- as usual, radian measure with the radius of the sphere being unity -- the area of the triangle is exactly equal to the sum of the angles minus tau/2.
Legendre uses the word "excess" to denote the triangle sum minus tau/2. In Euclidean geometry the excess of any triangle is zero, but in spherical geometry the excess of any triangle is its area. In fact, if we wanted to, we can use the words "excess" and "area" interchangeably in spherical geometry, as they are always equal.
Let's see how Legendre proves how excess equals area:
Demonstration. Let ABC be the triangle proposed; produce the sides until they meet the great circle DEFG drawn at pleasure without the triangle. By the preceding theorem the two triangles ADE, AGH , taken together, are equal to the lunary surface of which the angle is A, and which has for its measure 2A (495); thus we shall have ADE + AGH = 2A; for a similar reason BGF + BID = 2B, [and] CIH + CFE = 2C.
Oops -- Legendre calls the great circle DEFG, yet he throws in H and I in as well. To make the proof work, let's assume that our triangle lies in the Northern Hemisphere and the Equator is the great circle "drawn at pleasure." Then AB is extended to intersect the Equator at D and G (which are antipodal of course), AC intersects the Equator at E and H, and BC intersects the Equator at F and I.
"But the sum of these six triangles exceeds the surface of a hemisphere by twice the triangle ABC; moreover, the surface of a hemisphere is represented by 4 [right angles -- that is, 4lambda -- dw], consequently, the double of the triangle ABC is equal to 2A + 2B + 2C - 4[lambda], and consequently ABC = A + B + C - 2[lambda]; therefore every spherical triangle has for its measure the sum of its angles minus two right angles. QED"
(Notice that the six triangles in question cover ABC thrice and the rest of the Northern Hemisphere once, which is why Legendre writes that the six triangles cover the hemisphere plus twice ABC.)
We'll get to more consequences of the fact that excess equals area in my next spherical post. Next week, we'll continue with the next rule of my new middle school classroom.