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:

Area Postulate:

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*)

Proof:

Statements Reasons

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 2

*A*(495); thus we shall have

*ADE*+

*AGH*= 2

*A*; for a similar reason

*BGF*+

*BID*= 2

*B*, [and]

*CIH*+

*CFE*= 2

*C*.

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 2

*A*+ 2

*B*+ 2

*C*- 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.

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