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Chapter 6: Hyperbolic §6.1 Basic Theorems of .

MTH 411/511

Foundations of Geometry

MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 It’s good to have goals

Goals for today: • Introduce hyperbolic geometry. • Review basic theorems in hyperbolic geometry and properties of . • Prove the AAA condition.

MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 Universal Hyperbolic Theorem

First we recall a result from neutral geometry.

The Universal Hyperbolic Theorem (Theorem 4.9.1)

If there exists one `0, an external P0, and at least two lines that pass through P0 and are parallel to `0, then for every line ` and for every external point P there exist at least two lines that pass through P and are parallel to `.

(Corollary 4.9.2) The Universal Hyperbolic Theorem is equivalent to the negation of the Euclidean Parallel Postulate.

(Corollary 4.9.3) In any model for neutral geometry either the Euclidean Parallel Postulate or the Hyperbolic Parallel Postulate will hold.

MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 Hyperbolic geometry

In hyperbolic geometry we assume all of the axioms of neutral geometry and the Hyperbolic Parallel Postulate. Our goal in this section will be to develop enough of an understanding of Hyperbolic geometry so that we can outline a model (the Poincar´e Disk Model).

Hyperbolic Parallel Postulate For every line ` and for every point P that does not lie on `, there are at least two lines m and n such that P lies on both m and n and both m and n are parallel to `.

Hence, by (the corollary to) the Universal Hyperbolic Theorem, any statement equivalent to the Euclidean Parallel Postulate is false in hyperbolic geometry. Thus, the next results are immediate.

MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 Basic theorems in hyperbolic geometry

(Theorem 6.1.1) For every 4ABC, σ(4ABC) < 180◦.

(Corollary 6.1.2) For every triangle 4ABC, 0◦ < δ(4ABC) < 180◦.

(Theorem 6.1.3) ◦ For every convex ABCD, σ(ABCD) < 360 .

MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 Basic theorems in hyperbolic geometry

(Corollary 6.1.4) The summit in a Saccheri quadrilateral are acute.

(Corollary 6.1.4) The fourth angles in a Lamber quadrilateral is acute.

(Theorem 6.1.6) There does not exist a rectangle.

MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 Quadrilaterals

(Theorem 6.1.7) In a Lambert quadrilateral, the of a side between two right angles is strictly less than the length of the opposite side.

Proof. Let ABCD be a Lambert quadrilateral. Assume that the angles at vertices A, B, and C are right angles. We claim BC < AD. Since BC ≤ AD (Properties of Lambert quadrilaterals), it suﬃces to prove BC 6= AD.

Suppose BC = AD (RAA hypothesis). Then ABCD is a Saccheri quadrilateral so ∼ ∠BAD = ∠CDA (Properties of Saccheri quadrilaterals). Hence, ABCD is a rectangle, contradicting Theorem 6.1.6. So we reject the RAA hypothesis and conclude that BC 6= AD.

MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 Quadrilaterals

Deﬁnition 1

Let ABCD be a Saccheri quadrilateral with AB. The segment joining the of AB to the midpoint of CD is called the altitude of the quadrilateral. The length of the altitude is called the height of the quadrilateral.

By properties of Saccheri quadrilaterals, the altitude is to the base and the summit.

MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 Quadrilaterals

(Corollary 6.1.9) In a Saccheri quadrilateral, the length of the altitude is less than the length of a side.

(Corollary 6.1.10) In a Saccheri quadrilateral, the length of the summit is greater than the length of the base.

The next result proves that in hyperbolic geometry, implies congruence in .

MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 Similar/Congruent Triangles

AAA (Theorem 6.1.1) If 4ABC is similar to 4DEF , then 4ABC is congruent to 4DEF .

Proof. Let 4ABC and 4DEF be two triangles such that 4ABC ∼ 4DEF . We claim 4ABC =∼ 4DEF .

Suppose one pair of sides are congruent. Then 4ABC =∼ 4DEF (ASA). We now assume there are no pair of congruent sides (RAA hypothesis). WLOG, we assume AB > DE and AC > DF . −→ −→ Choose a point B0 on AB such that AB0 = DE and a point C 0 on AC such that 0 0 0 0 0 ∼ AC = DF (PCP). Then BCC B is convex (Theorem 4.6.7) and 4AB C = 4DEF 0 0 ∼ 0 0 ∼ (SAS). Thus ∠AB C = ∠ABC and ∠AC B = ∠ACB (deﬁnition of congruent 0 0 0 0 0 0 0 0 triangle). Since µ(∠CC B ) = 180 − µ(∠AC B ) and µ(∠BB C ) = 180 − µ(∠AB C ) 0 0 ◦ (LPT), then σ(BCC B ) = 360 , contradicting Theorem 6.1.3. Therefore we reject the RAA hypothesis and conclude that 4ABC =∼ 4DEF .

MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 Saccheri quadrilaterals

(Theorem 6.1.12) 0 0 0 0 If ABCD and A B C D are two Saccheri quadrilaterals such that 0 0 0 0 ∼ 0 0 ∼ 0 0 0 0 δ(ABCD) = δ(A B C D ) and CD = C D , then ABCD = A B C D .

Proof. 0 0 0 0 Let ABCD and A B C D be two Saccheri quadrilaterals such that 0 0 0 0 ∼ 0 0 ∼ 0 0 0 0 δ(ABCD) = δ(A B C D ) and CD = C D . We claim that ABCD = A B C D . −→ WLOG, suppose AD ≤ A0D0, so also BC ≤ B0C 0. Choose a point E on AD such that −→ DE = D0A” and a point F on CB such that CF = C 0B0. We have ∼ 0 0 0 ∼ ∼ 0 0 0 ∠ADC = ∠A D C = ∠DCB = D C B (Properties of Saccheri Triangles). Hence 4EDC =∼ 4A0D0C 0 (SAS). Since Saccheri quadrilaterals are convex, then ∼ 0 0 0 ∼ 0 0 0 ∠ECF = ∠A C B (AAP) so 4ECF = ∠A C B (SAS). Thus ∠EFC is a right ∼ 0 0 0 0 and a similar proof shows the same for ∠FED. It follows that EFCD = A B C D . ←→ ←→ Suppose E 6= A (RAA hypothesis). Then AB k EF (AIAT), so F 6= B and ABFE is a rectangle, contradicting Theorem 6.1.3. Thus we reject the RAA hypothesis and conclude that A = E and B = F .

MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020 Next time

Before next class: Read Section 6.2.

In the next lecture we will: • Discuss parallel and perpendicular lines in hyperbolic geometry. • Deﬁne common and use this to consider alternate interior angles.

MTH 411/511 (Geometry) Hyperbolic geometry Fall 2020