Download Discovery of Non-Euclidean Geometry

Discovery of Non-Euclidean Geometry
April 24, 2013
1
Hyperbolic geometry
János Bolyai (1802-1860), Carl Friedrich Gauss (1777-1855), and Nikolai Ivanovich Lobachevsky
(1792-1856) are three founders of non-Euclidean geometry.
Hyperbolic geometry is, by definition, the geometry that assume all the axioms for
neutral geometry and replace Hilbert’s parallel postulate by its negation, which is called the
hyperbolic axiom.
Hyperbolic axiom (Negation of Hilbert axiom). There exists a line l and a point P
not on l such that at least two distinct lines parallel to l pass through P .
Theorem 1.1. In hyperbolic geometry, all triangles have angle sum less than 180◦ , and all
convex quadrilaterals have angle sum less than 360◦ . In particular, there is no rectangle.
Proof. Trivial.
Theorem 1.2 (Universal Hyperbolic Theorem). In hyperbolic geometry, for every line l and
every point P not on l there pass through P at least two distinct lines parallel to l. In fact
there are infinitely many lines parallel to l through P .
t
t’
P
m
S
n
S’
Q
R
R’
n’
l
Figure 1: Existence of infinite parallels
Proof. Drop segment P Q perpendicular to l with foot Q on l. Erect line m at P perpendicular
to P Q. Then l, m are parallel. Pick a point R on l other than Q, and erect line t at R
perpendicular to l. Drop line n through P perpendicular t, intersecting t at S. If S is on
m, then S is the intersection m and t, and subsequently ¤P QRS is a rectangle, which is
impossible in hyperbolic geometry. So point S is not on m. Hence m, n are distinct lines
through P , both are parallel to l. See Figure 1.
Let R0 be point on l other than Q, R, and t0 be line through R0 perpendicular to l. There
exists line n0 through P perpendicular to t0 , intersecting t0 at S 0 . If P S = P S 0 , then ¤RR0 S 0 S
is a rectangle, which is impossible. So P S, P S 0 are distinct lines. Thus for all points R on l
other than Q, the lines P S through P perpendicular to l are all distinct.
1
Definition 1. Two triangles are ∆ABC and ∆A0 B 0 C 0 said to be similar if their vertices
can be put in on-to-one correspondence so that corresponding angles are congruent, i.e.,
∠A, ∠B, ∠C are congruent to angles ∠A0 , ∠B 0 , ∠C 0 respectively.
Theorem 1.3 (AAA criterion for congruence of hyperbolic triangles). In hyperbolic geometry, if two triangles are similar then they are congruent.
A
B
A’
C
B’’
B’
C’’
C’
Figure 2: Similar triangles are congruent in hyperbolic geometry
Proof. Given similar triangles ∆ABC and ∆A0 B 0 C 0 . Suppose the statement is not true, i.e.,
∆ABC is not congruent to ∆A0 B 0 C 0 . Then AB 6∼
= A0 B 0 , AC 6∼
= A0 C 0 , BC 6∼
= B 0 C 0 ; otherwise
∆ABC ∼
= ∆A0 B 0 C 0 by ASA. We may assume AB < A0 B 0 and AC < A0 C 0 . Lay off segment
AB on ray r(A0 , B 0 ) to have point B 00 such that AB ∼
= A0 B 00 and A0 ∗ B 00 ∗ B 0 , and lay off
0
0
00
segment AC on ray r(A , C ) to have point C such that AC ∼
= A0 C 00 and A0 ∗ C 00 ∗ C 0 . See
Figure 2. Then ∆ABC ∼
= ∆A0 B 00 C 00 by SAS. Hence
∠A0 B 00 C 00 ∼
= ∠B ∼
= ∠B 0 ,
∠A0 C 00 B 00 ∼
= ∠C ∼
= ∠C 0 .
It follows that lines B 00 C 00 , B 0 C 0 are parallel because of Congruent Corresponding Angles. So
¤B 0 C 0 C 00 B 00 is a convex quadrilateral. Since angles ∠A0 B 00 C 00 , ∠B 0 B 00 C 00 are supplementary,
angles ∠A0 C 00 B 00 , ∠C 0 C 00 B 00 are supplementary, and ∠B 0 ∼
= ∠A0 C 00 B 00 , then
= ∠A0 B 00 C 00 , ∠C 0 ∼
the angle sum of ¤B 0 C 0 C 00 B 00 is 360◦ . This is a contradiction.
Theorem 1.3 says that in hyperbolic geometry it is impossible to magnify or shrink
a triangle without distortion. So in hyperbolic world photography would be inherently
surrealistic.
Another consequence of Theorem 1.3 is that the length of a segment may be determined
by angles in hyperbolic geometry. For example, an angle of an equilateral triangle determines
the length of a side uniquely. This fact is sometimes referred to that hyperbolic geometry
has an absolute unit length.
2
Parallels that admit a common perpendicular
Given lines l, l0 and points A, B, C, . . . on l. Drop perpendiculars AA0 , BB 0 , CC 0 , . . . from
A, B, C, . . . to l0 with feet A0 , B 0 , C 0 , . . . on l0 respectively. We say that A, B, C, . . . are
equidistant from l0 if all these perpendicular segments are congruent to one another. See
Figure 3.
Theorem 2.1 (At most two points equidistant). Given two distinct parallels l, l0 in hyperbolic
geometry. Then any set of points on l equidistant from l0 contains at most two points.
Proof. Suppose it is not true, i.e., there is a set of three points A, B, C on l equidistant from
l0 . Then quadrilaterals ¤ABB 0 A0 , ¤ACC 0 A0 , ¤BCC 0 B 0 are Saccheri quadrilaterals (the base
2
A
B
C
A’
B’
C’
l
l’
Figure 3: No more than two equidistant points between two parallels
angles are right angles and the sides are congruent). Then the summit angles of the Saccheri
quadrilaterals are congruent, i.e.,
∠BAA0 ∼
= ∠ABB 0 ,
∠CAA0 ∼
= ∠ACC 0 ,
∠CBB 0 ∼
= ∠BCC 0 .
Thus ∠ABB 0 ∼
= ∠CBB 0 . Since ∠ABB 0 , ∠CBB 0 are supplementary, they must be right
angles. Hence all ¤ABB 0 A0 , ¤ACC 0 A0 , ¤BCC 0 B 0 are rectangles, which is impossible.
Lemma 2. Given a Saccheri quadrilateral ¤ABB 0 A0 with base right angles ∠A0 , ∠B 0 and
equal opposite sides AA0 , BB 0 . Let M, M 0 be the middle points of AB, A0 B 0 respectively.
Then segment M M 0 is perpendicular to both lines AB and A0 B 0 .
Proof. Draw segments A0 M and B 0 M . Note that AA0 ∼
= BB 0 , ∠A ∼
= ∠B, and AM ∼
= BM .
0
0
0
0
0
0 ∼
0
∼
∼
Then ∆A AM = ∆B BM by SAS. So A M = B M . Hence ∆A M M = ∆B M M 0 by
SSS. We then have ∠A0 M 0 M ∼
= ∠B 0 M 0 M . Subsequently, ∠A0 M 0 M and ∠B 0 M 0 M are right
angles. So M M 0 is perpendicular to the base A0 B 0 . Note that ∠A0 M M 0 ∼
= ∠B 0 M M 0 and
A
M
B
A’
M’
B’
Figure 4: Perpendicular middle point segment
∠AM A0 ∼
= ∠BM M 0 by angle addition. Subsequently, ∠AM M 0
= ∠BM B 0 . Then ∠AM M 0 ∼
0
and ∠BM M are right angles. So M M 0 is perpendicular to the summit AB.
Theorem 2.2 (Divergent and symmetric parallels). Let l, l0 be two lines perpendicular to a
segment M M 0 with M ∈ l, M 0 ∈ l0 .
(a) Then |M M 0 | < |XY 0 | for all X ∈ l, Y 0 ∈ l0 with XY 0 6= M M 0 .
(b) If M is the middle point of a segment AB on l, then A, B are equidistant from l0 .
(c) If M ∗ B ∗ D on l and BB 0 , DD0 are segments perpendicular to l0 with feet B 0 , D0 ∈ l0 ,
then BB 0 < DD0 .
Proof. (a) It is clear that M M 0 < M Y 0 for all Y 0 on l0 with Y 0 6= M 0 . Let X be a point on l
with X 6= M . Let XX 0 be segment perpendicular to l0 with foot X 0 on l0 . Then ¤M M 0 XX 0
is a Lambert quadrilateral. Thus
M M 0 < XX 0
by properties of Lambert quadrilaterals. Since XX 0 < XY 0 for Y 0 on l0 with Y 0 6= X 0 . We
see that M M 0 < XY 0 .
(b) Let AA0 , BB 0 be segments perpendicular to l0 with A0 , B 0 ∈ l0 . Draw segments AM 0
and BM 0 . Then ∆AM M 0 ∼
= ∆BM M 0 by SAS. So AM 0 ∼
= BM 0 and ∠AM 0 M ∼
= ∠BM 0 M .
3
C
C’
A
M
A’
M’
B
B’
l
D
l’
D’
Figure 5: Divergent parallels are symmetric
Subsequently, ∠AM 0 A0 ∼
= ∠BM 0 B 0 by angle subtraction. Thus ∆AA0 M 0 ∼
= ∆BB 0 M 0 by
0
0
0 ∼
0
0
0 ∼
SAA. Hence AA = BB and A M = B M .
(c) Note that ¤M 0 B 0 BM and ¤M 0 D0 DM are Lambert quadrilaterals. Then ∠B 0 BM and
∠D0 DM are acute angles. So ∠B 0 BD is obtuse, for it is supplementary to ∠B 0 BM . Hence
∠D0 DB = ∠D0 DM < ∠B 0 BD. Therefore BB 0 < DD0 by the property of ¤B 0 D0 DB.
Proposition 2.3 (Asymptotic and monotonic parallels). Given parallels l, l0 in hyperbolic
geometry, no two points of l are equidistant from l0 . Let AA0 , BB 0 , CC 0 be perpendicular
segments to l0 with A ∗ B ∗ C on l and A0 , B 0 , C 0 ∈ l0 . See Figure 6.
(a) If AA0 < BB 0 , then BB 0 < CC 0 .
(b) If BB 0 < CC 0 , then AA0 < BB 0 .
C
l
B
A
A’
B’
C’
l’
Figure 6: Monotone distance between asymptotic parallels
Proof. Consider quadrilaterals ¤A0 B 0 BA and ¤B 0 C 0 CB.
(a) Since AA0 < BB 0 , then ∠ABB 0 < ∠BAA0 . Since the angle sum of ¤A0 B 0 BA is less
than 360◦ , it follows that ∠ABB 0 is acute. So ∠CBB 0 is obtuse. Hence ∠BCC 0 must be
acute, since the angle sum of ¤B 0 C 0 CB is less than 360◦ . Of course ∠BCC 0 < ∠CBB 0 ,
subsequently, BB 0 < CC 0 by the property of ¤B 0 C 0 CB.
(b) Fix a point D on l with A ∗ B ∗ C ∗ D. For each X on the open ray r̊(D, A) we write
|DX| = x and define f (x) = |XX 0 |, where XX 0 is perpendicular to l0 with foot X 0 on l0 .
We claim that f (x) is a continuous function for x > 0. In fact, fix an x0 with point X0 on l
such that |BX0 | = x0 . Let X0 X00 be segment perpendicular to l0 with X00 ∈ l0 . Note that
|XX 0 | ≤ |XX00 | ≤ |X0 X00 | + |XX0 |,
|X0 X00 | ≤ |X0 X 0 | ≤ |XX 0 | + |XX0 |.
Then
½
|XX 0 | − |X0 X00 | if |XX 0 | ≥ |XX00 |
|X0 X00 | − |XX 0 | if |XX 0 | < |XX00 |
≤ |XX0 | = |x − x0 |.
|f (x) − f (x0 )| =
Clearly, f (x) is continuous at x0 . So f (x) is a continuous function for x > 0.
Suppose AA0 > BB 0 . Note that AA0 ∼
6= CC 0 . If |AA0 | < |CC 0 |, by intermediate value
theorem there exists a Y with B ∗ Y ∗ C such that |Y Y 0 | = |AA0 |. Then A, Y are equidistant
4
from l0 , which is impossible. If |AA0 | > |CC 0 |, by intermediate value theorem there exists a
point Z with A ∗ Z ∗ B such that |ZZ 0 | = |CC 0 |. Then C, Z are equidistant from l0 , which
is impossible.
We then must have AA0 < BB 0 .
3
Limiting parallel rays
Given a line l in hyperbolic geometry and a point P not on l. Let m be a line through P
parallel to l with left ray r(P, R). Drop perpendicular segment P Q to l with foot Q on l.
We consider rays between r(P, Q) and r(P, R), and want to find the critical ray r(P, X),
called the left limiting parallel ray to l through P , that does not meet l but any ray
between r(P, X) and r(P, Q) meets l. Likewise, there is a right limiting parallel ray to
l through P on the opposite side of P Q. See Figure 7.
Theorem 3.1. Given a line l and a point P not on l in hyperbolic geometry. Let P Q
be segment perpendicular to l with foot Q on l. Then there exist two non-opposite rays
r(P, X), r(P, X 0 ) on opposite sides of line P Q, satisfying the properties:
(a) Each of rays r(P, X), r(P, X 0 ) does not meet l.
(b) A ray r(P, Y ) meets l if and only if it is between r(P, X) and r(P, X 0 ).
(c) ∠QP X ∼
= ∠QP X 0 .
Proof. Let m be the line through P perpendicular to P Q. Pick a point R on the left side of
m and a point R0 on the right side of m separated by P . Draw segments QR and QR0 . Then
all rays between r(P, Q) and r(P, R) inclusive are represented by r(P, Y ) with Y ∈ QR. See
Figure 7.
R
P
R’
X
m
X’
U
T
S
V
V’
Q
l
Figure 7: Limiting parallel rays
(a) Let Σ1 be the set of points Y ∈ r(Q, R) such that the ray r(P, Y ) does not meet l,
and Σ2 the complement of Σ1 in QR. It is easy to see that both Σ1 , Σ2 are convex. So Σ1 , Σ2
form a Dedekind cut of QR. Then there exists a unique point X ∈ QR such that Σ1 , Σ2
are two rays (one of them is an open ray) of QR separated by X. We claim that X ∈ Σ1 .
Suppose X ∈ Σ2 , i.e., r(P, X) meets l at S. Pick a point T on l such that T ∗ S ∗ Q. Then
ray r(P, T ) is between r(P, R) and r(P, X). So r(P, T ) meets RQ at U and R ∗ U ∗ X, i.e.,
U ∈ Σ2 , which is a contradiction. The existence of ray r(P, X 0 ) is analogous.
(b) Since R ∈ Σ1 and Q ∈ Σ2 , we see that R ∗ X ∗ Q. It is obvious that if a ray r(P, Y )
is contained in the open half-plane opposite to H̊(m, Q) then r(P, Y ) does not meet l. We
then see that a ray r(P, Y ) meets l if and only if r(P, Y ) is between r(P, X) and r(P, X 0 ).
(c) Suppose that ∠XP Q is not congruent to ∠X 0 P Q, say, ∠XP Q < ∠X 0 P Q. Find point
0
V on l such that r(P, V 0 ) is between r(P, Q) and r(P, X 0 ), and ∠QP V 0 ∼
= ∠QP X. Mark
0
0
∼
∼
a point V on l such that V ∗ Q ∗ V and QV = QV . Then ∆P QV = ∆P QV 0 by SAS. So
∠QP V ∼
= ∠QP X , i.e., r(P, X) meets l at V , which is a contradiction.
= ∠QP V 0 ∼
5
The angle ∠QP X is called the angle of parallelism at point P with respect to l, its
degree measure is denoted Π(P Q)◦ . We have
Π(P Q)◦ < 90◦ .
4
Classification of parallels
Theorem 4.1. Given parallel lines l, l0 in hyperbolic geometry.
(a) If l contains a limiting parallel ray to l0 , then l, l0 are asymptotic parallels.
(b) If l does not contain limiting parallel ray to l0 , then l, l0 are divergent parallels.
D
l
C
R
P
m
B
X
A
S
T
A’
B’
Q
l’
C’
D’
Figure 8: The limiting parallel ray is asymptotic and monotonic
Proof. Fix a point P not on l0 and drop a perpendicular P Q to l0 with foot Q ∈ l0 . Let m
be the line through P perpendicular to P Q. Pick a point R on m other than P . Let r(P, X)
be a limiting parallel ray to l0 with X ∈ QR. See Figure 8.
(a) Let A, B, C, D be points on l with A ∗ B ∗ P ∗ C ∗ D and A, B ∈ r̊(P, X). Let
0
AA , BB 0 , CC 0 , DD0 be segments perpendicular to l0 with feet A0 , B 0 , C 0 , D0 ∈ l0 . Note that
∠XP Q is acute, ∠CP Q is obtuse, and the angle sum of ¤P QC 0 C is less than 360◦ . Then
∠P CC 0 is acute. Of course ∠P CC 0 < ∠CP Q. So P Q < CC 0 by property of quadrilaterals
with two base right angles.
Analogously, ∠P DD0 is acute and ∠DCC 0 is obtuse. Of course ∠P DD0 < ∠DCC 0 .
Then CC 0 < DD0 by property of quadrilaterals with two base right angles.
We claim |AA0 | ≤ |P Q| for all A on open ray r̊(P, X). Suppose |AA0 | > |P Q|. Let
S be a point on AA0 such that |AS| = 12 (|AA0 | − |P Q|). Clearly, |SA0 | > |P Q|. Then
∠A0 SP < ∠XP Q by property of quadrilateral with two base right angles. Of course ∠A0 SP
is acute. Since r(P, S) is between r(P, Q) and r(P, X), the ray r(P, S) meets l0 at T . Note
that ∠A0 ST is acute. So ∠A0 SP is obtuse, contradicting to that ∠A0 SP is acute.
We further claim AA0 < BB 0 for two points A, B on closed ray r(P, X) with A ∗ B ∗ C.
Suppose |AA0 | ≥ |BB 0 |. There exists a point E on BP (maybe B = P ) such that AA0 ∼
= EE 0
by continuity of distance function. Let M, M 0 be the middle points of AE, A0 E 0 respectively.
Then l, l0 are divergent parallels. Let F be on l such that F ∗ M ∗ C and M F ∼
= M C. We
have |F F 0 | = |CC 0 | > |P Q|, which is a contradiction.
(b) Assume that l does not contain any limiting parallel ray. If l = m, then l, l0 are
already divergent parallels. If l 6= m, we may assume that a ray r(P, Y ) of l is between
r(P, R) and r(P, X), where R ∗ Y ∗ X. It is easy to see that P Q < CC 0 < DD0 by similar
arguments.
Since ∠XP Y is acute, by Aristotle’s axiom there exists a point A on r(P, Y ) such that
AE > P Q, where AE is perpendicular to r(P, X) with foot E ∈ r(P, X). Of course AA0 >
6
P
R
A
D
C
l
m
Y
E
X
F
C’
Q
A’
l’
D’
Figure 9: Non-limiting parallel ray is symmetric
AF > AE. So AA0 > P Q. Thus l, l0 cannot be asymptotic (monotonic) parallels. So l, l0
must be divergent (symmetric) parallels.
Let A, A0 be two distinct points on the same side of a line AB such that lines AA0 , BB 0
are parallel. Then the figure, consisting of the segment AB (called the base) and the rays
r(A, A0 ) and r(B, B 0 ) (called the sides), is called a biangle with vertices A and B, denoted
∠A0 ABB 0 . See Figure 10. The interior of biangle ∠A0 ABB 0 is
˚
∠A0 AB ∩ ˚
∠ABB 0 .
∠A0 ABB 0 := ˚
If P ∈ ˚
∠A0 ABB 0 , either of rays r(A, P ), r(B, P ) is called an interior ray of ∠DABC. If
A’
A
D
Q
E
P
B
B’
G
F
Figure 10: Biangle and limiting parallel
each interior ray r(A, P ) intersects r(B, B 0 ), we say that r(A, A0 ) is limiting parallel to
r(B, B 0 ) and that biangle ∠A0 ABB 0 is closed at A, written r(A, A0 ) · r(B, B 0 ).
Lemma 3. Let ∠A0 ABB 0 be a biangle. See Figure 10.
(a) If D ∗ A ∗ A0 , then r(D, A0 ) · r(B, B 0 ) if and only if r(A, A0 ) · r(B, B 0 ).
(b) If r(A, A0 ) · r(B, B 0 ), so is r(B, B 0 ) · r(A, A0 ).
Proof. (a) Assume r(D, A0 ) · r(B, B 0 ). Take a point P in the interior of ∠A0 ABB 0 . It is
clear that P is an interior point of biangle ∠A0 DBB 0 . Then r(D, P ) meets r(B, B 0 ) at F
since ∠A0 DBB 0 is closed at D. Note that P is an interior point of ∠BAF . Then r(A, P ) is
between r(A, B) and r(A, F ). Thus r(A, P ) meets BF at G with B ∗ G ∗ C. By definition
r(A, A0 ) · r(B, B 0 ).
Conversely, assume r(A, A0 ) · r(B, B 0 ). For each ray r between r(D, B) and r(D, A0 ),
we have r meeting AB at E between A and B. Pick a point P on r such that D ∗E ∗P . Note
that ∠A0 AB > ∠AED ∼
= ∠BEP . There is a ray r(A, Q) such that ∠BAQ ∼
= ∠BEP . Then
0
r(A, Q)kr(E, P ). Since r(A, Q) meets r(B, B ), we see that r(E, P ) must meet r(B, B 0 ), i.e.,
r(D, P ) meet r(B, B 0 ). Hence ∠A0 DBB 0 is closed at D.
(b) Given an interior point P ∈ ∠ÅBB 0 and consider the ray r(B, P ). Suppose r(B, P )
does not meet r(A, A0 ). By the corollary of Aristotle’s axiom there exists a point Q on
r(B, P ) such that ∠AQB < ∠P BB 0 . See Figure 11. Note that r(A, Q) meets r(B, B 0 ) at
C. Then we have triangle ∆BCQ. Thus ∠AQB > ∠QBC = ∠P BB 0 , which contradicts
7
A’
A
Q
P
B
C
B’
Figure 11: Biangle and limiting parallel
∠AQB < ∠P BB 0 . So r(B, P ) must meet r(A, A0 ). Hence ∠A0 ABB 0 is closed at B, i.e.,
r(B, B 0 ) · r(A, A0 ).
Proposition 4.2 (Transitivity of limiting parallelism). If both rays r(A, A0 ) and r(B, B 0 )
are limiting parallel to ray r(C, C 0 ), then r(A, A0 ) and r(B, B 0 ) are limiting parallel to each
other.
A
A’
P
C
D
E
C’
E’
B
B’
F
Figure 12: Limiting parallels
Proof. Case 1. Lines AA0 and BB 0 are on opposite sides of line CC 0 . See Figure 12.
It is clear that r(A, A0 ) and r(B, B 0 ) have no point in common. Let AB meet CC 0 at D.
We may assume C ∗ D ∗ C 0 . Now for each point P interior to ∠A0 AB, the ray r(A, P ) meets
r(C, C 0 ) at E since r(A, A0 ) · r(C, C 0 ). We may assume C ∗ E ∗ C 0 . Then r(E, E 0 ) meets
r(B, B 0 ) at F since r(C, C 0 ) · r(B, B 0 ), where E ∗ E 0 ∗ F . Hence r(A, A0 ) meets r(B, B 0 ) at
F . Therefore by definition r(A, A0 ) and r(B, B 0 ) are limiting parallel to each other.
Case 2. Lines AA0 and BB 0 are on the same side of line CC 0 .
A
B’
D
B
D’
C
A’
C’
Figure 13: Limiting parallels
We first claim that AA0 and BB 0 do not meet. Suppose AA0 and BB 0 meet at point
D. We may assume that D belongs to both rays r(A, A0 ), r(B, B 0 ), and assume A ∗ D ∗ A0 ,
B ∗ D ∗ B 0 . Take a point D0 such that A ∗ D ∗ D0 . Then r(D, D0 ) meets r(C, C 0 ) since
r(D, B 0 ) · r(C, C 0 ), i.e., r(A, A0 ) meets r(C, C 0 ), which is a contradiction. See Figure 13.
Let AC meet BB 0 at point D. We may assume B ∗ D ∗ B 0 . For each point P interior
to ∠A0 AC, the ray r(A, P ) meets r(C, C 0 ) at point E. Since r(B, D) (= r(B, B 0 )) meets
the triangle ∆DCE, the ray r(B, B 0 ) meets either AE or CE. Since r(B, B 0 ) does not meet
r(C, C 0 ), so r(B, B 0 ) meet AE at F such that A ∗ F ∗ E. For point P interior to ∠BAD, the
ray r(A, P ) meets BD between B and D, of course r(A, P ) meets r(B, B 0 ). Hence r(A, A0 )
and r(B, B 0 ) are limiting parallel to each other.
8
A’
A
P
B
B’
F
D
C
C’
E
Figure 14: Limiting parallels
Two rays r, s are said to be limiting parallel, denoted r q s, if r ⊂ s or s ⊂ r or r · s.
Then q is an equivalence relation on rays in hyperbolic geometry. An equivalence class of rays
is called an ideal point or end, viewing it lying on each ray contained in the equivalence
class. Since a point on a line separates the line into two opposite rays, and opposite rays are
not equivalent, we see that every line has two ends on it.
If A, B are vertices of two rays r, s with r · s. Let R denote the ideal point determined
by these rays, i.e., R = [r] = [s]. We write r = AR and s = BR and refer to the closed
biangle with side r, s as a singly asymptotic triangle ∆ABR. We shall see that these
triangles have some properties in common with ordinary triangles.
Lemma 4. In hyperbolic geometry if two lines l, m are cut by a line t such that the alternate
interior angles are congruent, then l, m are divergent parallels.
t
Q’
P
Q
l
m
P’
Figure 15: Asymptotic triangle
Proof. Let t meet l at P and meet m at Q such that ∠Q0 P Q ∼
= ∠P 0 QP , where P P 0 is
perpendicular to m with foot P 0 on m and QQ0 is perpendicular to l with foot Q0 on l. Then
∆P QQ0 ∼
= QQ0 . Hence l, m are divergent parallel lines.
= ∆QP P 0 . So P P 0 ∼
Proposition 4.3. Let ∆ABR be a singly asymptotic triangle with a single ideal point R.
Then the exterior angles at A, B are greater than their respective opposite interior angles,
i.e., ∠A < ext ∠B.
A’
A
E
R
B’
B
D
C
Figure 16: Asymptotic triangle
9
Proof. Extend AB to C such that A ∗ B ∗ C. Draw ray r(B, D) such that ∠CBD ∼
= ∠BAA0
and extend DB to E such that E ∗ B ∗ D. Then ∠ABE ∼
= ∠CBD ∼
= ∠BAA0 . Thus
lines BD, AA0 are divergent parallels. Since r(B, B 0 ) · r(A, A0 ), we see that r(B, D) 6=
r(B, B 0 ). If r(B, D) is between r(B, B 0 ) and r(B, A), then r(B, D) meets r(A, A0 ), which is
a contradiction. So we must have r(B, D) between r(B, C) and r(B, B 0 ). This means that
∠CBD < ∠CBB 0 , i.e., ∠CBB 0 > ∠BAA0 .
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