Spherical Geometry (Legendre 481-486)

Last weekend marked the 50th anniversary of Singapore's independence. I only mention the holidays of other nations when there is some mathematical significance (such as last month's Bastille Day and its connection to the mathematician Galois.) The mathematical connection here is, of course, the popularity of Singapore math and desire for American math to be more like Singapore's. I've stated before that my opinion of Singapore math is mixed -- I like mentioning Singapore math to show that integrated math can still be rigorous, but the problem is that Singapore math may be too rigorous and sets up American students for failure.

Let's get back to spherical geometry and Legendre's Elements of Geometry. Today we are to cover six more propositions in Legendre. All six of these following theorems are provable in both Euclidean and spherical geometry, but they set us up well for the theorems that are unique to spherical geometry later on.

We begin with Legendre's Proposition 481:

481. Two triangles situated on the same sphere, or on equal spheres, are equal in all their parts, when a side and the two adjacent angles of one are equal to a side and the two adjacent angles of the other, each to each.

Demonstration. For one of these triangles may be applied to the other, as has been done in the analogous case of plane triangles (38).

Legendre's 480, if we recall, was SAS for spherical triangles, and we see that this proposition is in fact ASA for spherical triangles. He refers to his proposition 38 -- ASA for Euclidean triangles -- and points out that the two theorems may be proved in the same way.

As 480 was SAS and 481 was ASA, we can probably guess what 482 will be:

482. If two triangles on the same sphere, or on equal spheres, are equilateral with respect to each other, they will also be equiangular with respect to each other, and the equal angles will be opposite to equal sides.

Demonstration. This proposition is manifest from the reasoning pursuant in art. 478, by which it is shown that with three given sides AB, AC, BC, only two triangles can be constructed, differing as to the position of their parts, but equal as to the magnitude of these parts. Therefore two triangles, which are equilateral with respect to each other, are either absolutely equal, or at least equal by symmetry; in either case they are equiangular with respect to each other, and the equal angles are opposite to equal sides.

This theorem sounds a little awkward only because Legendre uses the word "equilateral" differently from the modern definition. Nowadays equilateral refers to one triangle whose three sides are congruent, but to Legendre, two triangles are "equilateral with respect to each other" if the sides of one are congruent to the corresponding sides of the other. So Proposition 482 is actually the analog of SSS -- so it goes along with SAS in 480 and ASA in 481. We notice that Legendre mentions his Proposition 478 and reminds us that two spherical triangles satisfying the SSS condition may either be "equal by symmetry" (opposite orientation) or "absolutely equal" (same orientation).

Now that we have completed the SAS, ASA, and SSS Congruence Theorems, you might be expecting the next propositions in Legendre to be something like AAS or HL. But as it turns out, neither AAS nor HL are valid in spherical geometry. The same counterexample disproves both of them -- we consider the triangles mentioned in my last spherical geometry post, each with a vertex at the North Pole and the other two on the Equator. We saw that in each triangle, the two angles with vertices on the Equator were both right angles, while the angles at the North Pole were not congruent, but supplementary (one has an angle of 62 degrees, the other 118 degrees).

We see that the AAS condition is satisfied -- each triangle has two right angles (A and A), as well as a side whose length equals a quadrant (S). We also see that the HL condition is satisfied -- each triangle is a right triangle with a side opposite to the right angle of quadrant length (H) and a side adjacent to the right triangle of quadrant length (L). And yet the two triangles are not congruent.

In Euclidean geometry, both AAS and HL are indirectly related to the theorems that are provable in both geometries. In particular, AAS is provable from ASA plus Triangle Angle-Sum Theorem. But in spherical geometry, we're still waiting for Legendre's Proposition 489 (which should appear in my next spherical geometry post) to tell us what the sum of the angles of a triangle is. We already know that it isn't 180 degrees. Therefore, the proof of AAS from ASA falls apart in spherical geometry.

Meanwhile, in Euclidean geometry, HL is provable from SSS plus the Pythagorean Theorem (though it isn't usually proved that way in most Geometry texts). But we can already see that the Pythagorean Theorem isn't valid in spherical geometry -- indeed, the hypotenuse of our counterexample above (to the extent that hypotenuse is defined as "side opposite the right angle") isn't even the longest side of the triangle, as both the hypotenuse and one of the other sides have the same length (a quadrant)! So instead of a^2 + b^2 = c^2, we have a^2 = c^2 (and therefore a^2 + b^2 > c^2). Therefore, the proof of HL from SSS falls apart in spherical geometry.

Let's go back to Legendre's Propositions. The rest of today's theorems actually will be valid in both Euclidean and spherical geometry.

483. In every isosceles spherical triangle the angles opposite to the equal sides are equal; and conversely, if two sides of a spherical triangle are equal, the triangle is isosceles.

Demonstration. 1. Let AB be equal to AC; we say that the angle C will be equal to the angle B. For, if from the vertex A the arc AD be drawn to the middle of the base, the two triangles ABD, ADC, will have the three sides of one equal to the three sides of the other, each to each, namely, AD common, BD = DC, AB = AC; consequently, by the preceding theorem, the two triangles will have their homologous angles equal, therefore B = C.

2. Let the angle B be equal to C; we say that AB will be equal to AC. For, if the side AB is not equal to AC, let AB be the greater; take BO = AC, and join OC. The two sides BO, BC, are equal to the two AC, BC; and the angle OBC contained by the first is equal to the angle ACB contained by the second. Consequently the two triangles have their other parts equal, namely OCB = ABC; but the angle ABC is, by hypothesis, equal to ACB; whence OCB is equal to ACB, which is impossible; AB then cannot be supposed unequal to AC; therefore the sides AB, AC, opposite to the equal angles B, C, are equal.

Legendre here is clearly proving the analog of the Isosceles Triangle Theorem. His use of the word "conversely" here implies that he will prove the Converse of the Isosceles Triangle Theorem as well.

Part 1 of his demonstration is on the forward Isosceles Triangle Theorem. We see that his proof can be converted easily into a two-column proof that is valid in both Euclidean and spherical geometry:

Given: AB = AC
Prove: Angle C = Angle B

Proof:
Statements                    Reasons
1. AB = AC                   1, Given
2. Draw D s.t. BD = DC 2. Ruler Postulate
3. AD = AD                   3. Reflexive Property of Congruence
4. Triangle ABD = ACD 4. SSS Congruence Theorem
5. Angle C = Angle B     5. CPCTC

Now Legendre's proof of the Converse of the Isosceles Triangle Theorem differs greatly from proofs given in most modern texts. Indeed, the U of Chicago text, in Section 7-3, proves the converse using AAS -- so this proof would be invalid in spherical geometry where we don't have AAS.

Instead, Legendre gives what appears, judging by his line "if the side AB is not equal to AC," to be an indirect proof (as what we want to prove is AB = AC). Without loss of generality, he assumes that AB > AC and lets O be the point that would make BO actually congruent to AC. He then shows that the triangles OCB and ABC would be congruent by SAS, and so the Angles OCB and ABC would be congruent by CPCTC. Then OCB would be congruent to ACB by the Transitive Property -- but this is a contradiction, as it would leave zero degrees for Angle ACO. So the assumption that AB and AC are not congruent is false, so AB and AC must be congruent. QED

I've mentioned before that mathematicians generally prefer proofs that generalize to more cases. From that perspective, Legendre's proof of Converse Isosceles is superior to the U of Chicago's, since the former is valid in both Euclidean and spherical geometry, while the latter is valid in only Euclidean but not spherical geometry. But in a high school classroom, we wish to avoid indirect proofs as long as possible. This is why teachers are more likely to use the AAS proof.

Incidentally, the U of Chicago proof is valid in "neutral geometry" -- because "neutral geometry" includes both Euclidean and hyperbolic, but not spherical geometry. In particular, AAS is valid in neutral geometry. But the proof of AAS given in most text books is not neutral because it uses the Triangle Angle-Sum Theorem of Euclidean geometry. Instead, there exists a proof of AAS that really is valid in neutral geometry, one that is more complicated than the ASA plus Triangle Sum proof. (As it turns out, it is another indirect proof that uses the Triangle Exterior Angle Inequality, or TEAI, that is valid in neutral geometry.)

So now we have two examples of theorems (Converse Isosceles and AAS) that are valid in geometries other than Euclidean, yet in Geometry texts are given proofs that aren't valid in all those geometries:

Converse Isosceles (theorem valid in neutral and spherical, proof valid in neutral only)
AAS (theorem valid in Euclidean and hyperbolic, proof valid in Euclidean only)

In each case, the proof that is valid in fewer geometries is a simpler direct proof, while the proof that is valid in more geometries is a trickier indirect proof. In high school classes, most texts usually prefer the simpler direct proof since they aren't actually teaching non-Euclidean geometry anyway.

Earlier, I wrote that I'm considering reorganizing my course so that as much geometry as possible is covered before any form of the Parallel Postulate (which would be the first topic of the second quarter of my course). This would entail that I would teach AAS before teaching the Parallel Postulate or Triangle Angle-Sum Theorems. But do I really want to make students learn indirect proofs so early in the year? Unless I feel that the students have a strong grasp on indirect proofs, I'm much more likely to save AAS for after the Parallel Postulate.

On the other hand, I said that my course would accommodate neutral geometry, but not necessarily spherical geometry. So I could still prove Converse Isosceles using AAS rather than the indirect proof given here in Legendre. Still, these are the sorts of decisions that I will have to make about my course once we reach the first day of school. Recall that many of these ideas go back to Dr. Franklin Mason, who tried to keep his geometry neutral as long as possible (including using the TEAI instead of Triangle Sun to prove neutral theorems), until he saw that the more complicated indirect proofs weren't necessarily relevant to PARCC or SBAC.

Let's get back to Legendre's next proposition, which he labels as a "scholium":

484. Scholium. It is evident, from the same demonstration, that the angle BAD = DAC, and the angle BDA = ADC. Consequently the two last are right angles; therefore, the arc drawn from the vertex of an isosceles spherical triangle to the middle of the base, is perpendicular to this base, and divides the angle opposite into two equal parts.

Legendre's "scholium" here appears as a theorem in Section 5-1 of the U of Chicago text, just before the Isosceles Triangle Theorem:

Theorem (from U of Chicago):
In an isosceles triangle, the bisector of the vertex angle, the perpendicular bisector of the base, and the median to the base determine the same line.

This is exactly what Legendre says in in Proposition 484 -- "the arc drawn from the vertex ... to the middle of the base" is the median, "is perpendicular to this base" is the perpendicular bisector, and "divides the angle opposite into two equal parts" is the angle bisector.

We proceed with Legendre's Proposition 485:

485. In any spherical triangle ABC, if the angle A is greater than the angle B, the side BC opposite to the angle A will be greater than the side AC opposite to the angle B; conversely, if the side BC is greater than AC, the angle A will be greater than the angle B.

Demonstration. 1. Let the angle A > B; make the angle BAD = B, and we shall have AD = DB (483); but
AD + DC > AC;
in the place of AD substitute DB, and we shall have DB + DC or BC > AC.

2. If we suppose BC > AC, we say that the angle BAC will be greater than ABC. For, if BAC were equal to ABC, we should have BC = AC, and if BAC were less than ABC, it would follow, according to what was just demonstrated, that BC < AC, which is contrary to the assumption; therefore the angle BAC is greater than ABC.

Proposition 485 is the analog of the Unequal Angles Theorem and Unequal Sides Theorem, given in Section 13-7 of the U of Chicago text. Once again, Legendre actually has two statements to prove.

The U of Chicago text begins by proving the Unequal Sides Theorem. But we notice that Step 3 of the U of Chicago's two-column proof gives the justification "Exterior Angle Inequality (with Triangle CC'A)" -- and we mentioned earlier that TEAI is not valid in spherical geometry. So Legendre cannot give the same proof as the U of Chicago does.

As it turns out, Legendre begins by proving the Unequal Angles Theorem first. His proof can be easily converted to two-column format, as follows:

Given: Triangle ABC with Angle A > B
Prove: BC > AC

Proof:
Statements                    Reasons
1. Angle A > B              1. Given
2. Ray AD inside BAC   2. Protractor Postulate
with Angle BAD = B
3. AD = DB                   3. Converse of the Isosceles Triangle Theorem
4. AD + DC > AC          4. Triangle Inequality
5. DB + DC > AC          5. Substitution (step 3 into step 4)
6. DB + DC = BC          6. Betweenness Theorem (Segment Addition Postulate)
7. BC > AC                   7. Substitution (step 6 into step 5)

If we compare this proof to the U of Chicago proof (of the Unequal Sides Theorem), we see that each step of this proof corresponds roughly with a step of the U of Chicago. The main difference is that the U of Chicago's TEAI (which is invalid in spherical geometry) has been replaced by the Triangle Inequality (which is valid in spherical geometry).

But there is actually a problem with this proof in my Geometry course given the way that my lessons are set up. We can't use Triangle Inequality to prove Unequal Angles because we actually needed Unequal Angles to prove the Triangle Inequality, thereby resulting in circularity.

Recall that the U of Chicago text gives Triangle Inequality as a postulate -- meaning that the U of Chicago could have given the Legendre proof without circularity. But if I wish to prove Triangle Inequality as a theorem, we can't use the Legendre proof.

Legendre gets from Unequal Angles to Unequal Sides almost the same way that the U of Chicago gets from Unequal Sides to Unequal Angles -- a sort of indirect proof in which two of three possibilities are ruled out.

Unlike the Converse of the Isosceles Triangle Theorem, proving Unequal Angles before Unequal Sides the way that Legendre does -- in the name of giving a proof that's valid in both Euclidean and spherical geometry -- wouldn't be too difficult for high school students. Also, notice that if we stick to Euclidean geometry, we can give the U of Chicago's proof of Unequal Sides and Legendre's proof of Unequal Angles (but with Triangle Inequality as a postulate) thus avoiding indirect proof altogether -- this may be ideal in a class where the students struggle with indirect proof.

Here is our final proposition for today:

486. If the two sides AB, AC, of the spherical triangle ABC are equal to the two sides DE, EF, of the triangle DEF described upon an equal sphere, if at the same time the angle A is greater than the angle D, we say that the third side BC of the first triangle will be greater than the third side EF of the second.

Notice that Legendre doesn't even bother to give a proof here -- instead he writes, "The demonstration of this proposition is entirely similar to that of art. 42." The theorem that we are discussing here is the analog of the SAS Inequality Theorem of Section 7-8, also known as the Hinge Theorem. Recall that the inequality theorems are spread out among several sections in the U of Chicago text. In my course, I'm considering giving all of the inequality theorems at once -- this allows us to give Triangle Inequality a proof, and SAS Inequality a proof based on the Triangle Inequality.

Legendre's proof of SAS Inequality, in his Proposition 42, is similar to the U of Chicago's. The key difference is that Legendre gives three cases, depending on whether the point the U of Chicago labels C' is inside, on, or outside the triangle. In the text, C' is outside the triangle.

Thus ends this post. This weekend I will provide the next "How to Fix Common Core" post, and so I hope that by next week I can post the big Propositions 487-489.