13 A Glimpse at Elliptic Geometry

13.1 Elliptic geometry is derived from spherical geometry

Problem 13.1. This is a quite interesting interpretation of the undefined terms ofincidence geometry:

A "point" is interpreted as a pair {P, Pa} of antipodal points on the unit sphere S2 inEuclidean three dimensional space.

A "line" is to be a great circle C on the sphere.

A "A point {P, Pa} lies on a line C" means that the great circle C containsboth antipodal points P and Pa.

Question. Which of the axioms of incidence geometry are satisfied, and which not?

Answer. All three axioms (I.1),(I.2) and (I.3) are satisfied.

Figure 13.1: A spherical line

Reason for (I.1). Given two ”points” {P, Pa} and {Q,Qa}, they determine in theambient Euclidean three space two different lines PPa and QQa, which both passthrough the origin (0, 0, 0). The two lines determine a plane , which intersects S2

in a great circle . This great circle is the ”line” connecting the two given points.

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Reason for (I.2). Given two ”points” {P, Pa} and {Q,Qa}, they determine the planeuniquely, and hence the great circle from above uniquely . Hence the ”line”connecting the two given points is unique.

Reason for (I.3). Every great circle contains infinitely many points, and hence morethan one antipodal pair of points. The three points (±1, 0, 0), (0,±1, 0) and(0, 0,±1) do not lie on a great circle.

Question. Which parallel property (elliptic, Euclidean, hyperbolic, or neither) doeshold?

Answer. The elliptic parallel property holds. Any two great circles intersect. Hencethere do not exist parallel lines.

13.2 The conformal model

Definition 13.1 (The stereographic model of elliptic geometry). Consider thefollowing model of elliptic geometry, built in the Euclidean plane as ambient underlyingreality. Throughout, we denote the open unit disk by

D = {(x, y) : x, y real, x2 + y2 < 1}

Its boundary is the unit circle

∂D = {(x, y) : x, y real, x2 + y2 = 1}

The center of D is denoted by O.

• The "points" are the points in the interior of the unit disk D as well as pairs ofantipodal points on its boundary ∂D.

• The "lines" are closed circular segments that pass through a pair of antipodeson ∂D.

Problem 13.2 (Construction of an elliptic line). Construct the elliptic line throughany two given points P and Q.

The solution uses pairs of antipodes and the Lemma below.

Definition 13.2 (The antipode). For any given point P , we define the antipode Pato be the point on the ray opposite to

−→OP such that OP · OPa = 1. For the center O

itself, the antipodal point is the point at infinity.

Lemma. If a circle passes through a pair of antipodes, it consists entirely of pairs ofantipodes. Hence a circle is an elliptic line if and only if it passes through a pair ofantipodes.

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Proof of the Lemma. Assume that a circle C passes through a pair of antipodes P andPa. Let Q be a third point on the circle. We draw the chords POPa and QO. Let Q2

be the second intersection point of the (Euclidean) line OQ with circle C. Because ofthe theorem of chords (Euclid III.35), we know that

OP · OPa = OQ · OQ2

By definition of antipodesOP · OPa = 1

and henceOQ · OQ2 = 1

The point O lies inside the circle C, and hence Q and Q2 lie on different sides of O, (Olies between Q and Q2.) This confirms that the second intersection point Q2 = Qa isthe antipode of Q .

Question. Provide the drawing for the Lemma.

Figure 13.2: An elliptic line consists of pairs of antipodes

Construction 13.1 (The antipode). To construct the antipode of a given point P ,one uses the theorems related to the Pythagorean Theorem. One erects the perpendicularh to radius OP at point O . Let C and Ca be the intersections of h with ∂D. Next oneerects the perpendicular on CP at point C. The antipodal point Pa is the intersectionat that perpendicular with the line OP .

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Indeed, by Thales’ theorem, the four points P,C, Pa and Ca lie on a circle. HenceEuclid III.35 implies that OP · OPa = OC · OCa = 1.

Figure 13.3: Construction of the antipode

Question. Provide the drawing for this construction.

Construction 13.2 (Elliptic line through two given points). Let two points Pand Q be given. To find the elliptic line through P and Q, we construct the antipode Paof one of the two points, say P . In the special case that P lies on ∂D, the constructionis easy, since PPa is a diameter of D. If P lies in the interior of D, and is not thecenter O itself, we proceed as explained above.

Finally, we need a circle C through the three points P, Pa and Q. Its center is theintersection of the perpendicular bisectors of segments PQ, PPa and QPa.

Reason for validity of the construction. Thus the circle C consists entirely of pairs ofantipodes. Hence the two circles C and ∂D intersect in the pair of antipodes A and Aa.Hence C is an elliptic line.

Problem 13.3. Do the construction for an example. Put in—as a check—the commonchord of the two circles. It should come out to be a diameter of ∂D.

13.3 Falsehood of the exterior angle theorem

Proposition 13.1 (No Exterior Angle Theorem). The Exterior Angle Theoremdoes not hold in spherical geometry. Neither does it hold in elliptic geometry.

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Figure 13.4: Construction of an elliptic line

The following construction gives a counterexample. For simplicity, take antipodes B andD = Ba on the equator ∂D. We connect them by two different half great circles a andc, lying symmetric to the south pole O. Next draw a segment through O that cuts cat point A and a at point C, but not at a right angle. Nevertheless, you get a figurewhich is point symmetric by point O. Hence the triangle �ABC has the interior angleγ = ∠ACB congruent to the nonadjacent exterior angle δ = ∠DAC.

Question. Draw the figure, as explained, in the conformal model, just the southernhemisphere. Measure and report the three angles of your �ABC. What do you observeabout segments AB and CD, and what about segments BC and AD.

Answer. The segments AB ∼= CD are congruent, and segments BC ∼= AD are congru-ent.

Proposition 13.2 (No SAA Congruence). SAA congruence does not hold in spher-ical geometry. Neither does it hold in elliptic geometry.

Proof. The following construction gives an example to confirm this claim. Choose anyangle γ < 45◦. We construct two non-congruent triangles with AB = 45◦, α = 90◦ andγ as given.

Take two pairs of antipodes A,Aa and B,Ba on the equator with AB = 45◦. Wechose for C any point on the diameter AOAa, thus getting a right angle α = 90◦. LetC2 be the point such that segment CC2 has midpoint O. Finally, one needs to get thegreat semi-circles BCBa and BC2Ba. They are point symmetric with respect to centerO.

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Figure 13.5: The exterior angle can be congruent or smaller than a nonadjacent interiorangle.

Question (a). Draw the figure as explained into the southern hemisphere of the confor-mal model. Stress the triangles �ACB and �AC2B with two different colors.

Question (b). Measure and report the angles γ = ∠ACB and γ2 = ∠AC2B.

Answer.

Question (c). Which pieces of the two triangles are pairwise congruent?

Answer. Obviously the common side AB is congruent to itself. Both triangles have aright angle at vertex A. Because of point symmetry of the entire figure by O, they arecongruent. Angles γ1 = ∠ACB ∼= γ2 = ∠AC2B are congruent, too.

Question (d). For which congruence theorem are these just the matched pieces? Nev-ertheless, the two triangles �ACB and �AC2B are not congruent!

Answer. The matching pieces are those required for SAA congruence (Theorem 25 inHilbert’s Foundations).

Question (e). What do you observe about the two (corresponding but not congruent)sides of triangles �ACB and �AC2B, which lie across the second given angle, and werenot prescribed initially?

Answer. These are the two sides BC and BC2. They turn out to be supplementary.

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Figure 13.6: Two non congruent triangles matching in two angles and the side across oneof them.

Figure 13.7: SAA congruence in neutral geometry

13.4 Area of a spherical triangle

Proposition 13.3 (The area of a spherical triangle). The area of a sphericaltriangle is proportional to the excess of its sum of angles over two right angles. With itsangles measured in radians, the area of a spherical �ABC on a sphere of radius R is

A = (α + β + γ − π)R2

By a lune LA is denoted the region lying between two half great circle.

Question (a). Let α be the angle at vertex A between the two great half circles boundingthe lune, measured in radians. Find a formula for the area the area |LA| of this lune.

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Figure 13.8: Another example of two non congruent triangles matching in two angles andthe side across one of them.

Answer. The area |LA| is proportional to α. For α = π, one gets a half sphere whichhas area 2π R2. Hence the area of the lune between to half great circle intersecting atangle α is

|LA| = 2αR2

Now we want to find the area of �ABC. To get an easy drawing, I use the conformalmodel. We can assume that vertices A and B lie on the equator ∂D, and vertex C onthe southern hemisphere, I mean inside disk D.

Question (b). Draw an example. I find the case with acute angles α and β easier todraw. Extend the side AC to the antipode Aa on the equator ∂D, and furthermore tothe antipode Ca on the northern hemisphere outside the disk D. Similarly, extend theside BC to the antipode Ba on the equator ∂D, and furthermore to the antipode Ca

on the northern hemisphere, outside the disk D. Check whether the three points C,Ca

and O lie on a Euclidean line.

In the figure, we use three lunes:

(a) the lune LA with tips A and Aa, bounded by the extended sides ABAa and ACAa

of �ABC.

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Figure 13.9: A spherical triangle

(b) the lune LB with tips B and Ba, bounded by the extended sides BABa and BCBa

of �ABC.

(c) the lune LC with tips C and Ca, bounded by the extended sides CAaCa and CBaCa.

The first two lunes lie totally on the southern hemisphere (interior of disk D). Only thethird lune wraps around the equator to the northern hemisphere (exterior of disk D).

Question (c). You may color the three lunes with three different colors. (I have useddifferent textures instead only for photocopying.) What are the areas of the three lunes?

Answer.

(1) |LA| = 2αR2 , |LB| = 2β R2 , |LC | = 2αR2

We now derive two equations for areas. The two lunes LA and LB overlap in the�ABC. They cover the entire southern hemisphere except of the �CAaBa. Since ahemisphere has area 2π R2, one gets for the areas

(2) |LA| + |LB| = |LA ∩ LB| + |LA ∪ LB| = |�ABC| + 2π R2 − |�CAaBa|

The lune LC consists just of the two non-overlapping triangles �AaBaCa and �CAaBa.Hence

(3) |LC | = |�CAaBa| + |�AaBaCa|

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Question (d). Add the two equations (2)(3) and use (1). Get the area of �ABC.

Answer. Adding (2) and (3) yields

|LA| + |LB| + |LC | = |�ABC| + 2π R2 − |�CAaBa| + |�CAaBa| + |�AaBaCa|= |�ABC| + 2π R2 + |�AaBaCa|= 2|�ABC| + 2π R2

Because the areas of the lunes are |LA| = 2αR2 , |LB| = 2β R2 , |LC| = 2αR2, wefinally get

|�ABC| = (α + β + γ − π)R2

13.5 Does Pythagoras’ imply the parallel postulate?

A purely geometric proof would have to struggle with some obvious difficulties. At firstthe squares of the sides of a triangle cannot be interpreted as area of any figure in neitherspherical nor hyperbolic geometry. In these geometries, there do not exist any similarfigures other than congruent ones, and squares do not exist at all. So the squares of thesides can only be interpreted as numbers. I am for now considering these difficulties asnot crucial. Here are possible approaches to address the question posed:

(1) One can look at the special Pythagorean triples of integers such that a2 + b2 = c2.The smallest one is the triple 32 + 42 = 52. So a meaningful question is whetherthere exist right triangles with the sides 3, 4 and 5 in either spherical or hyperbolicgeometry.

(2) Take any right triangle �ABC. In spherical geometry its sides satisfy

cos c = cos a cos b

Does this imply that Pythagoras’ Theorem can never hold?

(3) Take any right triangle �ABC. In hyperbolic geometry—with Gaussian curvatureK = −1—the sides of a right triangle satisfy

cosh c = cosh a cosh b

Does this imply that Pythagoras’ Theorem can never hold?

I do no know about any approach to answer at least part of these questions in theframework of neutral geometry—where they would naturally fit, similarly to Legendre’sTheorems about the angle sum. My approach problems (2) and (3) uses substantialtools from trigonometry and calculus.

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13.5.1 Main results obtained by calculus

Theorem 13.1. If a spherical right triangle has the short arc as its hypothenuse c, thisside is strictly shorter than for the Euclidean flat triangle with legs of the same lengths.

Hence in single elliptic geometry, the hypothenuse c of any right triangle is strictlyshorter than for the corresponding Euclidean flat triangle with legs of the same lengths.

Theorem 13.2 (”The lunes of Pythagoras”). For spherical right triangles wherethe hypothenuse c is the longer arc, this side may be either shorter or longer, or of equallength as for the Euclidean flat triangle with legs of the same lengths.

Assume the length of the hypothenuse is restricted to the interval c ∈ (π, 2π). Thelegs satisfy a2 + b2 = c2—as in the Euclidean case— if and only if (a, b) lies on onecurve C ⊂ E, which connects the boundary points (s∗, 0) and (0, s∗) inside the quartercircle

E = {(a, b) : 0 < a < 2π , 0 < b < 2π , a2 + b2 < 4π2}Here the number s∗ ≈ 257◦ is the unique solution of tan s∗ = s∗ ∈ (π, 3π

2).

Theorem 13.3. In hyperbolic geometry, the hypothenuse c is longer than for the Eu-clidean flat triangle with legs of the same lengths.

13.5.2 Examples for the ”lunes of Pythagoras”

Definition 13.3 (”The lunes of Pythagoras”). The lunes of Pythagoras are rightspherical triangles, the sides a, b, c ∈ (0, 2π) of which satisfy both

(13.1)cos c = cos a cos b and

c2 = a2 + b2

On the pages 596 through 599, one can see several examples—obtained by solvingthe equation

cos√a2 + x2 − cos a cos x

x2:= g(a, x) = 0

numerically for several values of a. I show the flat triangle, its sides in radian measure,together with the spherical triangle in stereographic projection, with side b on the equa-tor (red), and side a extended through the south pole (blue). Sides c and b intersect inboth vertex A and its antipode. The lengths of the sides are given in degree measure.

Problem 13.4. Give your own comments about these figures.

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Figure 13.10: The Pythagoras lune with a = 45◦.

Figure 13.11: The Pythagoras lune with a = 70◦.

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Figure 13.12: The Pythagoras lune with a = 90◦ is highly degenerate.

Figure 13.13: The Pythagoras lune with a = 120◦.

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Figure 13.14: The Pythagoras lune with a = 150◦.

Figure 13.15: The Pythagoras lune with a = 180◦ corresponds to a flat right triangle withsides 3, 4 and 5—it is degenerate, too.

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Figure 13.16: An approximately isosceles lune.

13.6 The stereographic projection

Definition 13.4 (The stereographic projection). In Euclidean 3-space, take asphere S2 of diameter 1. The stereographic projection is the central projection whichmaps S2 onto the tangent plane T to the sphere at the south pole S, with the northpole N as center of projection. The north pole N itself is mapped to a single point ∞at infinity.

The equator is mapped to the unit circle ∂D. The southern hemisphere is mappedto the disk D, and the northern hemisphere is mapped to the exterior R2 ∪ {∞} \Dof the disk.

Main Theorem 23 (Properties of the stereographic projection).

(1) Circles are mapped to circles or Euclidean lines.

(2) Angles are preserved.

13.6.1 The stereographic projection is an inversion

To master this tricky three-dimensional problem, it is helpful to use three-dimensionalinversion, defined by the straightforward generalization of definition 13.5. Thus thepreservation of circles by the stereographic projections can be derived from the preser-vation of circles by inversion.

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Let B be the open interior of a three dimensional ball of radius R and center Z, anddenote its boundary sphere by ∂B.

Definition 13.5 (Inversion by a sphere). The inversion by the sphere ∂B is definedto be the mapping from the Euclidean 3-space plus one point ∞ at infinity to itself,which maps an arbitrary point P �= Z, to its inverse point P ′—defined to be the point

on the ray−→OP such that |OP | · |OP ′| = R2. Hence, especially, all the points of ∂B are

mapped to themselves. The inversion maps the origin Z to ∞, and ∞ to Z.

Proposition 13.4 (The stereographic projection as an inversion). Let ∂B be thesphere with center Z = N at the north pole through the south pole S. Since the principalsphere S2 has diameter 1, the new sphere ∂B has radius 1.

Inversion by ∂B maps the sphere S2 to the the tangent plane T at the south poleS. The stereographic projection is the restriction of the inversion by ∂B to the principalsphere S2.

Proof. Again the drawing plane is determined by the meridian S1 through the northpole N , south pole S and the generic point A. Let ∂B be the intersection of ∂B withthe drawing plane—which is the circle around the north pole N through the south poleS. Inversion by ∂B maps the north pole N to infinity, and the south pole S to itself.The diameter NS is mapped to itself. The circle with diameter NS and the tangent t atthe south pole are both orthogonal to this diameter NS. Hence preservation of anglesimplies that the circle with diameter NS is mapped to the tangent t.

Rotation of the entire figure around the axis NS yields a three dimensional figure,from which one sees that inversion by ∂B maps the sphere S2 to the the tangent planeT at the south pole S. This mapping takes any point A ∈ S2 along the projection ray−−→NA to the image point P ∈ T . Hence it is just the stereographic projection.

Direct check using triangle geometry. Take any generic point A on the median S1 andlet P be the image of A by the stereographic projection. From the similar right triangles�NSA and �NPS, one gets the proportion

|NA||NS| =

|NS||NP | = cos(∠ANS)

and hence|NA| · |NP | = |NS|2 = 1

which confirms that point P is the inverse image of A by a circle of radius 1 aroundnorth pole N , as to be shown.

Reason for item (1) of the Main Theorem. We show that the image of any circle on thesphere S2 is a circle or line in the tangent plane T through the south pole S.

Take any circle C on the sphere S2. Denote its center by A. In the special casethat the center A is north pole N or south pole S, one can immediately show that C

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Figure 13.17: The stereographic projections maps circles to circles.

is mapped to a circle. Hence we exclude this special case, and let B and C be the twopoints of the circle lying on the meridian NA. Suppose the stereographic projectionsmaps the points A,B and C to the image points P,Q and R in the tangent plane T .

We use the plane through the north pole N , south pole S and point A as our drawingplane. Points B,C, P,Q,R lie in this plane, too.

Let U be the circle through the points B,C and Q. Because the pairs A,P , B,Qand C,R are inverse points by circle ∂B,

|NB| · |NQ| = |NC| · |NR| = 1

Now Euclid’s theorem of chords implies that circle U consists entirely of pairs of inversepoints. Hence all four points B,C,Q,R lies on circle U . Furthermore circle U is orthog-onal to ∂B. Because of the orthogonality, inversion by the circle ∂B maps the circle Uto itself.

Rotation around the common diameter of the orthogonal circles U ⊥ ∂B producesthe two orthogonal spheres U ⊥ ∂B. Again because of the orthogonality, inversion bythe sphere ∂B— which is just the stereographic projection—maps U to itself.

Since the stereographic projection maps the sphere S2 to the tangent plane T , itmaps the intersection S2 ∩ U to the intersection U ∩ T . But U ∩ S2 clearly is a circle.Because the centers of both spheres U and S2 lie in the drawing plane, this circle hasdiameter BC, and hence U ∩ S2 = C is the originally given circle. Its stereographicimage is U ∩T , which is a circle in the tangent plane T , as to be shown. Again, becausethe center of sphere U lies in the drawing plane, the image circle U ∩ T has diameterQR.

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Reason for item (2) of the Main Theorem. Let any three points A,B,C on the spherebe mapped to the points P,Q,R on the tangent plane t. The two lines PQ and PRhave as preimages two circles ABN and ACN on the sphere S2.

It is enough to show that the angle ∠QPR between the two lines is congruent to theangle between the two circles. We can measure the angle between these two circles asthe angle between their tangent at the second intersection point N .

The tangent t1 to the circle ABN at the north pole N is a line in the plane spanned

by the projection rays−−→NA and

−−→NB. It is the parallel to line PQ through the north

pole N . Similarly, the tangent t2 to the circle ACN at the north pole N is the parallelto line PR through the north pole N . Because parallel shift leaves angles invariant, theangle between the lines PQ and PR is congruent to the angle between the tangents t1and t2, as to be shown.

13.6.2 Many congruent angles

Figure 13.18: Antipodes on the sphere are mapped to antipodes in the conformal model.

Proposition 13.5 (Antipodes).

(1) Antipodes A and Aa on S2 are mapped to antipodal points P and Pa with respect tothe circle ∂D.

(2) Great circles (elliptic lines) are mapped to circles or Euclidean lines through a pairof antipodes.

(3) Points on the northern and southern hemispheres of S2 which are symmetrical tothe equatorial plan, are mapped to inverse points with respect to the circle ∂D.

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Reason for item (1). Draw a section in the plane with points N,S and the generic pointA on the sphere S2. Let P be the image of A. The points O,P,Aa, Pa lie in the sameplane. Angle ∠ANAa = 90◦ by Thales’ theorem, and angle ∠NSP = 90◦, because theradius is orthogonal to the tangent by Euclid III.16. The altitude theorem for the righttriangle �PaNP yields

|SP | · |SPa| = |NS|2 = 1

Hence P and Pa are antipodal points with respect to the unit circle ∂D.

Figure 13.19: Many congruent angles

Let S1 be the meridian through point A. This is just the great circle lying in ourdrawing plane. Let a be the tangent line to the circle S1 at point A. Let R be theintersection point of that tangent with line t = SP .

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Let β be the congruent base angles of the isosceles triangle �NOA. We claim thattriangle �RAP is isosceles, too. Indeed, angle addition at vertex A shows that

β + 90◦ + ∠RAP = 180◦

and the angle sum of the triangle �NPS yields

β + 90◦ + ∠RPA = 180◦

Hence the triangle �RAP has congruent base angles

α = ∠RAP ∼= ∠RPA

and hence it is isosceles.

Question. Here is a drawing of what has been explained so far. In this figure, find and markwith same color angle α = 90◦ − β at vertices N,P, Pa, A,Aa five times. Find and markwith another color of your choice angle β at these vertices as often as possible. (I donot take into account vertical angles.)

Figure 13.20: Points symmetric to the equatorial plane are mapped to inverse points withrespect to the equator circle ∂D.

Reason for item (3). Take any generic point B on the median S1 and let Q be the imageby the stereographic projection. The antipode Ba has the image point Qa such that

|SQ| · |SQa| = 1

and the south pole S lies between Q and Qa. Rotating the antipode Ba by 180◦ aroundthe axis NS yields the point C, such that B and C are reflection images by the equatorial

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plane. The stereographic projection maps point C to point R, which is obtained frompoint Qa by a rotation of 180◦ around the south pole S. Hence

|SQ| · |SR| = |SQ| · |SQa| = 1

and the south pole S lies outside the segment QR. Clearly this confirms that R is theinversion image of Q by the equator circle ∂D.

Direct check of item (3) with triangle geometry. Take any generic point B on the me-dian S1 and let Q be the image by the stereographic projection. From the similar righttriangles �NSB and �NQS, one gets the proportion

|QS||NS| =

|BS||NB| = tan(∠BNS)

Now let point C be the reflection image of B by the equatorial plan, and R be thestereographic image of C. From the similar right triangles �NSC and �NRS, onegets the proportion

|RS||NS| =

|CS||NC| = tan(∠CNS)

Since segments BC and NS are parallel, the triangles �NSB ∼= �SNC are congruent.Hence |BS| = |NC| and |NB| = |SC|, and finally

|QS||NS| ·

|RS||NS| =

|BS||NB| ·

|CS||NC| = 1

and hence|QS| · |RS| = |NS|2 = 1

which confirms that the stereographic images Q and R of the symmetric points B andC are inverse images by the equator circle ∂D.

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Figure 13.21: The diameter in the drawing plane is projected symmetrically.

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