Chapter 4: Neutral geometry §4.8 Rectangles and defect §4.9 The Universal Hyperbolic Theorem

MTH 411/511

Foundations of Geometry

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Goals for today: • Define Saccheri and Lambert quadrilaterals. • Study properties of these quadrilaterals. • State and prove the Universal Hyperbolic Theorem.

It’s good to have goals

MTH 411/511 (Geometry) Neutral geometry Fall 2020 • Define Saccheri and Lambert quadrilaterals. • Study properties of these quadrilaterals. • State and prove the Universal Hyperbolic Theorem.

It’s good to have goals

Goals for today:

MTH 411/511 (Geometry) Neutral geometry Fall 2020 • Study properties of these quadrilaterals. • State and prove the Universal Hyperbolic Theorem.

It’s good to have goals

Goals for today: • Define Saccheri and Lambert quadrilaterals.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 • State and prove the Universal Hyperbolic Theorem.

It’s good to have goals

Goals for today: • Define Saccheri and Lambert quadrilaterals. • Study properties of these quadrilaterals.

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

Goals for today: • Define Saccheri and Lambert quadrilaterals. • Study properties of these quadrilaterals. • State and prove the Universal Hyperbolic Theorem.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 (Clairaut’s Axiom) There exists a rectangle.

(Corollary 4.8.7) Clairaut’s Axiom is equivalent to the Euclidean Parallel Postulate.

However, other similar objects exist in ubiquity in neutral geometry.

Rectangles

In the previous lecture we considered existence of rectangles and proved that they only exist in models where the Euclidean Parallel Postulate holds.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 (Corollary 4.8.7) Clairaut’s Axiom is equivalent to the Euclidean Parallel Postulate.

However, other similar objects exist in ubiquity in neutral geometry.

Rectangles

In the previous lecture we considered existence of rectangles and proved that they only exist in models where the Euclidean Parallel Postulate holds. (Clairaut’s Axiom) There exists a rectangle.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 However, other similar objects exist in ubiquity in neutral geometry.

Rectangles

In the previous lecture we considered existence of rectangles and proved that they only exist in models where the Euclidean Parallel Postulate holds. (Clairaut’s Axiom) There exists a rectangle.

(Corollary 4.8.7) Clairaut’s Axiom is equivalent to the Euclidean Parallel Postulate.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Rectangles

In the previous lecture we considered existence of rectangles and proved that they only exist in models where the Euclidean Parallel Postulate holds. (Clairaut’s Axiom) There exists a rectangle.

(Corollary 4.8.7) Clairaut’s Axiom is equivalent to the Euclidean Parallel Postulate.

However, other similar objects exist in ubiquity in neutral geometry.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 The segment AB is called the base and CD the summit. The two right angles are the base angles and the other two the summit angles.

Saccheri quadrilaterals

Definition 1 A Saccheri quadrilateral is a quadrilateral ABCD such that ∠ABC and ∠DAB are right angles and AD =∼ BC.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 The two right angles are the base angles and the other two the summit angles.

Saccheri quadrilaterals

Definition 1 A Saccheri quadrilateral is a quadrilateral ABCD such that ∠ABC and ∠DAB are right angles and AD =∼ BC. The segment AB is called the base and CD the summit.

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

Definition 1 A Saccheri quadrilateral is a quadrilateral ABCD such that ∠ABC and ∠DAB are right angles and AD =∼ BC. The segment AB is called the base and CD the summit. The two right angles are the base angles and the other two the summit angles.

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

Definition 1 A Saccheri quadrilateral is a quadrilateral ABCD such that ∠ABC and ∠DAB are right angles and AD =∼ BC. The segment AB is called the base and CD the summit. The two right angles are the base angles and the other two the summit angles.

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

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 1. the diagonals AC and BD are congruent 2. the summit angles ∠BCD and ∠ADC are congruent, 3. the segment joining the midpoint of AB to the midpoint of CD is perpendicular to both AB and CD, 4. ABCD is a parallelogram, 5. ABCD is a convex quadrilateral, and 6. the summit angles ∠BCD and ∠ADC are either right or acute.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Proof. ∼ ∼ (1) Consider the triangles 4ACB and 4BDA. Since AD = CB and ∠DAB = ∠CBA (definition of Saccheri quadrilaterals), then 4ACB =∼ 4BDA (SAS). Thus, AC =∼ BD (definition of congruent triangles).

(2) Consider the triangles 4ADC and 4BCD. By (1), AC =∼ BD. Since AD =∼ BC (definition of Saccheri quadrilaterals), then 4ADC =∼ 4BCD (SSS). Thus, ∼ ∠ADC = ∠BCD (definition of congruent triangles).

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 1. the diagonals AC and BD are congruent, 2. the summit angles ∠BCD and ∠ADC are congruent,

MTH 411/511 (Geometry) Neutral geometry Fall 2020 ∼ ∼ Since AD = CB and ∠DAB = ∠CBA (definition of Saccheri quadrilaterals), then 4ACB =∼ 4BDA (SAS). Thus, AC =∼ BD (definition of congruent triangles).

(2) Consider the triangles 4ADC and 4BCD. By (1), AC =∼ BD. Since AD =∼ BC (definition of Saccheri quadrilaterals), then 4ADC =∼ 4BCD (SSS). Thus, ∼ ∠ADC = ∠BCD (definition of congruent triangles).

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 1. the diagonals AC and BD are congruent, 2. the summit angles ∠BCD and ∠ADC are congruent,

Proof. (1) Consider the triangles 4ACB and 4BDA.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 then 4ACB =∼ 4BDA (SAS). Thus, AC =∼ BD (definition of congruent triangles).

(2) Consider the triangles 4ADC and 4BCD. By (1), AC =∼ BD. Since AD =∼ BC (definition of Saccheri quadrilaterals), then 4ADC =∼ 4BCD (SSS). Thus, ∼ ∠ADC = ∠BCD (definition of congruent triangles).

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 1. the diagonals AC and BD are congruent, 2. the summit angles ∠BCD and ∠ADC are congruent,

Proof. ∼ ∼ (1) Consider the triangles 4ACB and 4BDA. Since AD = CB and ∠DAB = ∠CBA (definition of Saccheri quadrilaterals),

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Thus, AC =∼ BD (definition of congruent triangles).

(2) Consider the triangles 4ADC and 4BCD. By (1), AC =∼ BD. Since AD =∼ BC (definition of Saccheri quadrilaterals), then 4ADC =∼ 4BCD (SSS). Thus, ∼ ∠ADC = ∠BCD (definition of congruent triangles).

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 1. the diagonals AC and BD are congruent, 2. the summit angles ∠BCD and ∠ADC are congruent,

Proof. ∼ ∼ (1) Consider the triangles 4ACB and 4BDA. Since AD = CB and ∠DAB = ∠CBA (definition of Saccheri quadrilaterals), then 4ACB =∼ 4BDA (SAS).

MTH 411/511 (Geometry) Neutral geometry Fall 2020 (2) Consider the triangles 4ADC and 4BCD. By (1), AC =∼ BD. Since AD =∼ BC (definition of Saccheri quadrilaterals), then 4ADC =∼ 4BCD (SSS). Thus, ∼ ∠ADC = ∠BCD (definition of congruent triangles).

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 1. the diagonals AC and BD are congruent, 2. the summit angles ∠BCD and ∠ADC are congruent,

Proof. ∼ ∼ (1) Consider the triangles 4ACB and 4BDA. Since AD = CB and ∠DAB = ∠CBA (definition of Saccheri quadrilaterals), then 4ACB =∼ 4BDA (SAS). Thus, AC =∼ BD (definition of congruent triangles).

MTH 411/511 (Geometry) Neutral geometry Fall 2020 By (1), AC =∼ BD. Since AD =∼ BC (definition of Saccheri quadrilaterals), then 4ADC =∼ 4BCD (SSS). Thus, ∼ ∠ADC = ∠BCD (definition of congruent triangles).

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 1. the diagonals AC and BD are congruent, 2. the summit angles ∠BCD and ∠ADC are congruent,

Proof. ∼ ∼ (1) Consider the triangles 4ACB and 4BDA. Since AD = CB and ∠DAB = ∠CBA (definition of Saccheri quadrilaterals), then 4ACB =∼ 4BDA (SAS). Thus, AC =∼ BD (definition of congruent triangles).

(2) Consider the triangles 4ADC and 4BCD.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Since AD =∼ BC (definition of Saccheri quadrilaterals), then 4ADC =∼ 4BCD (SSS). Thus, ∼ ∠ADC = ∠BCD (definition of congruent triangles).

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 1. the diagonals AC and BD are congruent, 2. the summit angles ∠BCD and ∠ADC are congruent,

Proof. ∼ ∼ (1) Consider the triangles 4ACB and 4BDA. Since AD = CB and ∠DAB = ∠CBA (definition of Saccheri quadrilaterals), then 4ACB =∼ 4BDA (SAS). Thus, AC =∼ BD (definition of congruent triangles).

(2) Consider the triangles 4ADC and 4BCD. By (1), AC =∼ BD.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Thus, ∼ ∠ADC = ∠BCD (definition of congruent triangles).

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 1. the diagonals AC and BD are congruent, 2. the summit angles ∠BCD and ∠ADC are congruent,

Proof. ∼ ∼ (1) Consider the triangles 4ACB and 4BDA. Since AD = CB and ∠DAB = ∠CBA (definition of Saccheri quadrilaterals), then 4ACB =∼ 4BDA (SAS). Thus, AC =∼ BD (definition of congruent triangles).

(2) Consider the triangles 4ADC and 4BCD. By (1), AC =∼ BD. Since AD =∼ BC (definition of Saccheri quadrilaterals), then 4ADC =∼ 4BCD (SSS).

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

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 1. the diagonals AC and BD are congruent, 2. the summit angles ∠BCD and ∠ADC are congruent,

Proof. ∼ ∼ (1) Consider the triangles 4ACB and 4BDA. Since AD = CB and ∠DAB = ∠CBA (definition of Saccheri quadrilaterals), then 4ACB =∼ 4BDA (SAS). Thus, AC =∼ BD (definition of congruent triangles).

(2) Consider the triangles 4ADC and 4BCD. By (1), AC =∼ BD. Since AD =∼ BC (definition of Saccheri quadrilaterals), then 4ADC =∼ 4BCD (SSS). Thus, ∼ ∠ADC = ∠BCD (definition of congruent triangles).

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Proof. (3) Let M be the midpoint of AB and N the midpoint of CD (Existence of Midpoints). Then CN =∼ DN (definition of the midpoint). Thus, 4AND =∼ 4BNC (part (2) and SSS). Now AN =∼ BN (definition of congruent triangles). Since AM =∼ BM (definition ∼ ∼ of the midpoint) then 4ANM = 4BNM (SSS). It follows that ∠AMN = ∠BMN and they form a linear pair. Thus, they are right angles and one can show similarly that ∠DNM and ∠CNM are right angles.

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 3. the segment joining the midpoint of AB to the midpoint of CD is perpendicular to both AB and CD,

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Then CN =∼ DN (definition of the midpoint). Thus, 4AND =∼ 4BNC (part (2) and SSS). Now AN =∼ BN (definition of congruent triangles). Since AM =∼ BM (definition ∼ ∼ of the midpoint) then 4ANM = 4BNM (SSS). It follows that ∠AMN = ∠BMN and they form a linear pair. Thus, they are right angles and one can show similarly that ∠DNM and ∠CNM are right angles.

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 3. the segment joining the midpoint of AB to the midpoint of CD is perpendicular to both AB and CD,

Proof. (3) Let M be the midpoint of AB and N the midpoint of CD (Existence of Midpoints).

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Thus, 4AND =∼ 4BNC (part (2) and SSS). Now AN =∼ BN (definition of congruent triangles). Since AM =∼ BM (definition ∼ ∼ of the midpoint) then 4ANM = 4BNM (SSS). It follows that ∠AMN = ∠BMN and they form a linear pair. Thus, they are right angles and one can show similarly that ∠DNM and ∠CNM are right angles.

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 3. the segment joining the midpoint of AB to the midpoint of CD is perpendicular to both AB and CD,

Proof. (3) Let M be the midpoint of AB and N the midpoint of CD (Existence of Midpoints). Then CN =∼ DN (definition of the midpoint).

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Now AN =∼ BN (definition of congruent triangles). Since AM =∼ BM (definition ∼ ∼ of the midpoint) then 4ANM = 4BNM (SSS). It follows that ∠AMN = ∠BMN and they form a linear pair. Thus, they are right angles and one can show similarly that ∠DNM and ∠CNM are right angles.

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 3. the segment joining the midpoint of AB to the midpoint of CD is perpendicular to both AB and CD,

Proof. (3) Let M be the midpoint of AB and N the midpoint of CD (Existence of Midpoints). Then CN =∼ DN (definition of the midpoint). Thus, 4AND =∼ 4BNC (part (2) and SSS).

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Since AM =∼ BM (definition ∼ ∼ of the midpoint) then 4ANM = 4BNM (SSS). It follows that ∠AMN = ∠BMN and they form a linear pair. Thus, they are right angles and one can show similarly that ∠DNM and ∠CNM are right angles.

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 3. the segment joining the midpoint of AB to the midpoint of CD is perpendicular to both AB and CD,

Proof. (3) Let M be the midpoint of AB and N the midpoint of CD (Existence of Midpoints). Then CN =∼ DN (definition of the midpoint). Thus, 4AND =∼ 4BNC (part (2) and SSS). Now AN =∼ BN (definition of congruent triangles).

MTH 411/511 (Geometry) Neutral geometry Fall 2020 ∼ ∼ then 4ANM = 4BNM (SSS). It follows that ∠AMN = ∠BMN and they form a linear pair. Thus, they are right angles and one can show similarly that ∠DNM and ∠CNM are right angles.

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 3. the segment joining the midpoint of AB to the midpoint of CD is perpendicular to both AB and CD,

Proof. (3) Let M be the midpoint of AB and N the midpoint of CD (Existence of Midpoints). Then CN =∼ DN (definition of the midpoint). Thus, 4AND =∼ 4BNC (part (2) and SSS). Now AN =∼ BN (definition of congruent triangles). Since AM =∼ BM (definition of the midpoint)

MTH 411/511 (Geometry) Neutral geometry Fall 2020 ∼ It follows that ∠AMN = ∠BMN and they form a linear pair. Thus, they are right angles and one can show similarly that ∠DNM and ∠CNM are right angles.

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 3. the segment joining the midpoint of AB to the midpoint of CD is perpendicular to both AB and CD,

Proof. (3) Let M be the midpoint of AB and N the midpoint of CD (Existence of Midpoints). Then CN =∼ DN (definition of the midpoint). Thus, 4AND =∼ 4BNC (part (2) and SSS). Now AN =∼ BN (definition of congruent triangles). Since AM =∼ BM (definition of the midpoint) then 4ANM =∼ 4BNM (SSS).

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Thus, they are right angles and one can show similarly that ∠DNM and ∠CNM are right angles.

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 3. the segment joining the midpoint of AB to the midpoint of CD is perpendicular to both AB and CD,

Proof. (3) Let M be the midpoint of AB and N the midpoint of CD (Existence of Midpoints). Then CN =∼ DN (definition of the midpoint). Thus, 4AND =∼ 4BNC (part (2) and SSS). Now AN =∼ BN (definition of congruent triangles). Since AM =∼ BM (definition ∼ ∼ of the midpoint) then 4ANM = 4BNM (SSS). It follows that ∠AMN = ∠BMN and they form a linear pair.

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

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 3. the segment joining the midpoint of AB to the midpoint of CD is perpendicular to both AB and CD,

Proof. (3) Let M be the midpoint of AB and N the midpoint of CD (Existence of Midpoints). Then CN =∼ DN (definition of the midpoint). Thus, 4AND =∼ 4BNC (part (2) and SSS). Now AN =∼ BN (definition of congruent triangles). Since AM =∼ BM (definition ∼ ∼ of the midpoint) then 4ANM = 4BNM (SSS). It follows that ∠AMN = ∠BMN and they form a linear pair. Thus, they are right angles and one can show similarly that ∠DNM and ∠CNM are right angles.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Proof. ←→ (4) Let A0 be a point on AD such that A0 ∗ A ∗ D and A0 6= A (Ruler Postulate). Then A0AB is a right angle and angles A0AB and ABC are a pair of alternate interior ∠ ←→ ←→ ∠ ∠←→ angles for the lines AD and BC with transversal AB (definition of Saccheri ←→ ←→ ←→ ←→ quadrilaterals). Thus, AD k BC (AIAT). Similarly, AB k DC.

(5) We proved on a previous homework assignment that parallelograms are convex.

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 4. ABCD is a parallelogram, 5. ABCD is a convex quadrilateral, and

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Then A0AB is a right angle and angles A0AB and ABC are a pair of alternate interior ∠ ←→ ←→ ∠ ∠←→ angles for the lines AD and BC with transversal AB (definition of Saccheri ←→ ←→ ←→ ←→ quadrilaterals). Thus, AD k BC (AIAT). Similarly, AB k DC.

(5) We proved on a previous homework assignment that parallelograms are convex.

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 4. ABCD is a parallelogram, 5. ABCD is a convex quadrilateral, and

Proof. ←→ (4) Let A0 be a point on AD such that A0 ∗ A ∗ D and A0 6= A (Ruler Postulate).

MTH 411/511 (Geometry) Neutral geometry Fall 2020 ←→ ←→ ←→ ←→ Thus, AD k BC (AIAT). Similarly, AB k DC.

(5) We proved on a previous homework assignment that parallelograms are convex.

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 4. ABCD is a parallelogram, 5. ABCD is a convex quadrilateral, and

Proof. ←→ (4) Let A0 be a point on AD such that A0 ∗ A ∗ D and A0 6= A (Ruler Postulate). Then A0AB is a right angle and angles A0AB and ABC are a pair of alternate interior ∠ ←→ ←→ ∠ ∠←→ angles for the lines AD and BC with transversal AB (definition of Saccheri quadrilaterals).

MTH 411/511 (Geometry) Neutral geometry Fall 2020 ←→ ←→ Similarly, AB k DC.

(5) We proved on a previous homework assignment that parallelograms are convex.

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 4. ABCD is a parallelogram, 5. ABCD is a convex quadrilateral, and

Proof. ←→ (4) Let A0 be a point on AD such that A0 ∗ A ∗ D and A0 6= A (Ruler Postulate). Then A0AB is a right angle and angles A0AB and ABC are a pair of alternate interior ∠ ←→ ←→ ∠ ∠←→ angles for the lines AD and BC with transversal AB (definition of Saccheri ←→ ←→ quadrilaterals). Thus, AD k BC (AIAT).

MTH 411/511 (Geometry) Neutral geometry Fall 2020 (5) We proved on a previous homework assignment that parallelograms are convex.

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 4. ABCD is a parallelogram, 5. ABCD is a convex quadrilateral, and

Proof. ←→ (4) Let A0 be a point on AD such that A0 ∗ A ∗ D and A0 6= A (Ruler Postulate). Then A0AB is a right angle and angles A0AB and ABC are a pair of alternate interior ∠ ←→ ←→ ∠ ∠←→ angles for the lines AD and BC with transversal AB (definition of Saccheri ←→ ←→ ←→ ←→ quadrilaterals). Thus, AD k BC (AIAT). Similarly, AB k DC.

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

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 4. ABCD is a parallelogram, 5. ABCD is a convex quadrilateral, and

Proof. ←→ (4) Let A0 be a point on AD such that A0 ∗ A ∗ D and A0 6= A (Ruler Postulate). Then A0AB is a right angle and angles A0AB and ABC are a pair of alternate interior ∠ ←→ ←→ ∠ ∠←→ angles for the lines AD and BC with transversal AB (definition of Saccheri ←→ ←→ ←→ ←→ quadrilaterals). Thus, AD k BC (AIAT). Similarly, AB k DC.

(5) We proved on a previous homework assignment that parallelograms are convex.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Proof. ∼ (6) By (2), ∠BCD = ∠ADC. Set x = µ(∠BCD) = µ(∠ADC). Because ABCD is convex,

◦ 360 ≥ σ(ABCD) = µ(∠ABC) + µ(∠BCD) + µ(∠CDA) + µ(∠DAB) = 90◦ + x + x + 90◦.

Thus, 180◦ ≥ 2x, so x ≤ 90◦.

Saccheri quadrilaterals

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 6. the summit angles ∠BCD and ∠ADC are either right or acute

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

Properties of Saccheri quadrilaterals (Theorem 4.8.10)

If ABCD is a Saccheri quadrilateral with base AB, then 6. the summit angles ∠BCD and ∠ADC are either right or acute

Proof. ∼ (6) By (2), ∠BCD = ∠ADC. Set x = µ(∠BCD) = µ(∠ADC). Because ABCD is convex,

◦ 360 ≥ σ(ABCD) = µ(∠ABC) + µ(∠BCD) + µ(∠CDA) + µ(∠DAB) = 90◦ + x + x + 90◦.

Thus, 180◦ ≥ 2x, so x ≤ 90◦.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Lambert quadrilaterals

Definition 2 A Lambert quadrilateral is a quadrilateral in which three of the angles are right angles.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Lambert quadrilaterals

Definition 2 A Lambert quadrilateral is a quadrilateral in which three of the angles are right angles.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Proof. (1) This is similar to the proof of part (4) in the previous theorem.

(2) This follows because ABCD is a parallelogram by part (1). (3) This is similar to part (6) in the previous theorem.

Lambert quadrilaterals

Properties of Lambert quadrilaterals (Theorem 4.8.11) If ABCD is a Lambert quadrilateral with right angles are vertices A, B and C, then 1. ABCD is a parallelogram, 2. ABCD is a convex quadrilateral, 3. angle ∠ADC is either right or acute, and 4. BC ≤ AD.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 (2) This follows because ABCD is a parallelogram by part (1). (3) This is similar to part (6) in the previous theorem.

Lambert quadrilaterals

Properties of Lambert quadrilaterals (Theorem 4.8.11) If ABCD is a Lambert quadrilateral with right angles are vertices A, B and C, then 1. ABCD is a parallelogram, 2. ABCD is a convex quadrilateral, 3. angle ∠ADC is either right or acute, and 4. BC ≤ AD.

Proof. (1) This is similar to the proof of part (4) in the previous theorem.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 (3) This is similar to part (6) in the previous theorem.

Lambert quadrilaterals

Properties of Lambert quadrilaterals (Theorem 4.8.11) If ABCD is a Lambert quadrilateral with right angles are vertices A, B and C, then 1. ABCD is a parallelogram, 2. ABCD is a convex quadrilateral, 3. angle ∠ADC is either right or acute, and 4. BC ≤ AD.

Proof. (1) This is similar to the proof of part (4) in the previous theorem.

(2) This follows because ABCD is a parallelogram by part (1).

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Lambert quadrilaterals

Properties of Lambert quadrilaterals (Theorem 4.8.11) If ABCD is a Lambert quadrilateral with right angles are vertices A, B and C, then 1. ABCD is a parallelogram, 2. ABCD is a convex quadrilateral, 3. angle ∠ADC is either right or acute, and 4. BC ≤ AD.

Proof. (1) This is similar to the proof of part (4) in the previous theorem.

(2) This follows because ABCD is a parallelogram by part (1). (3) This is similar to part (6) in the previous theorem.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Proof. −→ (4) Suppose AD < BC (RAA hypothesis). Choose a point D0 on AD such that 0 0 0 0 AD = BC (PCP), so D 6= D . Hence, AD CB is a Saccheri quadrilateral, so ∠D CB is either right or acute (Properties of Saccheri quadrilaterals (6)). Since D is between −→ −−→ −→ A and D0, then CD is between CD0 and CB. Thus,

0 0 0 90 ≥ µ(∠D CB) = µ(∠D CD) + µ(∠DCB) = µ(∠D CD) + 90

0 (ACP). Then 0 ≥ µ(∠D CD), a contradiction. Thus, we reject the RAA hypothesis and conclude that AD ≥ BC.

Lambert quadrilaterals

Properties of Lambert quadrilaterals (Theorem 4.8.11) If ABCD is a Lambert quadrilateral with right angles are vertices A, B and C, then 1. ABCD is a parallelogram, 2. ABCD is a convex quadrilateral, 3. angle ∠ADC is either right or acute, and 4. BC ≤ AD.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 −→ Choose a point D0 on AD such that 0 0 0 0 AD = BC (PCP), so D 6= D . Hence, AD CB is a Saccheri quadrilateral, so ∠D CB is either right or acute (Properties of Saccheri quadrilaterals (6)). Since D is between −→ −−→ −→ A and D0, then CD is between CD0 and CB. Thus,

0 0 0 90 ≥ µ(∠D CB) = µ(∠D CD) + µ(∠DCB) = µ(∠D CD) + 90

0 (ACP). Then 0 ≥ µ(∠D CD), a contradiction. Thus, we reject the RAA hypothesis and conclude that AD ≥ BC.

Lambert quadrilaterals

Properties of Lambert quadrilaterals (Theorem 4.8.11) If ABCD is a Lambert quadrilateral with right angles are vertices A, B and C, then 1. ABCD is a parallelogram, 2. ABCD is a convex quadrilateral, 3. angle ∠ADC is either right or acute, and 4. BC ≤ AD.

Proof. (4) Suppose AD < BC (RAA hypothesis).

MTH 411/511 (Geometry) Neutral geometry Fall 2020 0 0 Hence, AD CB is a Saccheri quadrilateral, so ∠D CB is either right or acute (Properties of Saccheri quadrilaterals (6)). Since D is between −→ −−→ −→ A and D0, then CD is between CD0 and CB. Thus,

0 0 0 90 ≥ µ(∠D CB) = µ(∠D CD) + µ(∠DCB) = µ(∠D CD) + 90

0 (ACP). Then 0 ≥ µ(∠D CD), a contradiction. Thus, we reject the RAA hypothesis and conclude that AD ≥ BC.

Lambert quadrilaterals

Properties of Lambert quadrilaterals (Theorem 4.8.11) If ABCD is a Lambert quadrilateral with right angles are vertices A, B and C, then 1. ABCD is a parallelogram, 2. ABCD is a convex quadrilateral, 3. angle ∠ADC is either right or acute, and 4. BC ≤ AD.

Proof. −→ (4) Suppose AD < BC (RAA hypothesis). Choose a point D0 on AD such that AD0 = BC (PCP), so D 6= D0.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Since D is between −→ −−→ −→ A and D0, then CD is between CD0 and CB. Thus,

0 0 0 90 ≥ µ(∠D CB) = µ(∠D CD) + µ(∠DCB) = µ(∠D CD) + 90

0 (ACP). Then 0 ≥ µ(∠D CD), a contradiction. Thus, we reject the RAA hypothesis and conclude that AD ≥ BC.

Lambert quadrilaterals

Properties of Lambert quadrilaterals (Theorem 4.8.11) If ABCD is a Lambert quadrilateral with right angles are vertices A, B and C, then 1. ABCD is a parallelogram, 2. ABCD is a convex quadrilateral, 3. angle ∠ADC is either right or acute, and 4. BC ≤ AD.

Proof. −→ (4) Suppose AD < BC (RAA hypothesis). Choose a point D0 on AD such that 0 0 0 0 AD = BC (PCP), so D 6= D . Hence, AD CB is a Saccheri quadrilateral, so ∠D CB is either right or acute (Properties of Saccheri quadrilaterals (6)).

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Thus,

0 0 0 90 ≥ µ(∠D CB) = µ(∠D CD) + µ(∠DCB) = µ(∠D CD) + 90

0 (ACP). Then 0 ≥ µ(∠D CD), a contradiction. Thus, we reject the RAA hypothesis and conclude that AD ≥ BC.

Lambert quadrilaterals

Properties of Lambert quadrilaterals (Theorem 4.8.11) If ABCD is a Lambert quadrilateral with right angles are vertices A, B and C, then 1. ABCD is a parallelogram, 2. ABCD is a convex quadrilateral, 3. angle ∠ADC is either right or acute, and 4. BC ≤ AD.

Proof. −→ (4) Suppose AD < BC (RAA hypothesis). Choose a point D0 on AD such that 0 0 0 0 AD = BC (PCP), so D 6= D . Hence, AD CB is a Saccheri quadrilateral, so ∠D CB is either right or acute (Properties of Saccheri quadrilaterals (6)). Since D is between −→ −−→ −→ A and D0, then CD is between CD0 and CB.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 0 Then 0 ≥ µ(∠D CD), a contradiction. Thus, we reject the RAA hypothesis and conclude that AD ≥ BC.

Lambert quadrilaterals

Properties of Lambert quadrilaterals (Theorem 4.8.11) If ABCD is a Lambert quadrilateral with right angles are vertices A, B and C, then 1. ABCD is a parallelogram, 2. ABCD is a convex quadrilateral, 3. angle ∠ADC is either right or acute, and 4. BC ≤ AD.

Proof. −→ (4) Suppose AD < BC (RAA hypothesis). Choose a point D0 on AD such that 0 0 0 0 AD = BC (PCP), so D 6= D . Hence, AD CB is a Saccheri quadrilateral, so ∠D CB is either right or acute (Properties of Saccheri quadrilaterals (6)). Since D is between −→ −−→ −→ A and D0, then CD is between CD0 and CB. Thus,

0 0 0 90 ≥ µ(∠D CB) = µ(∠D CD) + µ(∠DCB) = µ(∠D CD) + 90 (ACP).

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Thus, we reject the RAA hypothesis and conclude that AD ≥ BC.

Lambert quadrilaterals

Properties of Lambert quadrilaterals (Theorem 4.8.11) If ABCD is a Lambert quadrilateral with right angles are vertices A, B and C, then 1. ABCD is a parallelogram, 2. ABCD is a convex quadrilateral, 3. angle ∠ADC is either right or acute, and 4. BC ≤ AD.

Proof. −→ (4) Suppose AD < BC (RAA hypothesis). Choose a point D0 on AD such that 0 0 0 0 AD = BC (PCP), so D 6= D . Hence, AD CB is a Saccheri quadrilateral, so ∠D CB is either right or acute (Properties of Saccheri quadrilaterals (6)). Since D is between −→ −−→ −→ A and D0, then CD is between CD0 and CB. Thus,

0 0 0 90 ≥ µ(∠D CB) = µ(∠D CD) + µ(∠DCB) = µ(∠D CD) + 90

0 (ACP). Then 0 ≥ µ(∠D CD), a contradiction.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Lambert quadrilaterals

Properties of Lambert quadrilaterals (Theorem 4.8.11) If ABCD is a Lambert quadrilateral with right angles are vertices A, B and C, then 1. ABCD is a parallelogram, 2. ABCD is a convex quadrilateral, 3. angle ∠ADC is either right or acute, and 4. BC ≤ AD.

Proof. −→ (4) Suppose AD < BC (RAA hypothesis). Choose a point D0 on AD such that 0 0 0 0 AD = BC (PCP), so D 6= D . Hence, AD CB is a Saccheri quadrilateral, so ∠D CB is either right or acute (Properties of Saccheri quadrilaterals (6)). Since D is between −→ −−→ −→ A and D0, then CD is between CD0 and CB. Thus,

0 0 0 90 ≥ µ(∠D CB) = µ(∠D CD) + µ(∠DCB) = µ(∠D CD) + 90

0 (ACP). Then 0 ≥ µ(∠D CD), a contradiction. Thus, we reject the RAA hypothesis and conclude that AD ≥ BC.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 That is, as the next theorem asserts, uniqueness of a parallel for one line and external point guarantees unique parallels for all lines and all possible external points.

The Universal Hyperbolic Theorem (Theorem 4.9.1)

If there exists one line `0, an external point 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.

The Universal Hyperbolic Theorem

Just as existence of triangles with defect 0◦ is all or nothing in neutral geometry, so two are unique parallels.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 The Universal Hyperbolic Theorem (Theorem 4.9.1)

If there exists one line `0, an external point 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.

The Universal Hyperbolic Theorem

Just as existence of triangles with defect 0◦ is all or nothing in neutral geometry, so two are unique parallels. That is, as the next theorem asserts, uniqueness of a parallel for one line and external point guarantees unique parallels for all lines and all possible external points.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 (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.

The Universal Hyperbolic Theorem

Just as existence of triangles with defect 0◦ is all or nothing in neutral geometry, so two are unique parallels. That is, as the next theorem asserts, uniqueness of a parallel for one line and external point guarantees unique parallels for all lines and all possible external points.

The Universal Hyperbolic Theorem (Theorem 4.9.1)

If there exists one line `0, an external point 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 `.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 (Corollary 4.9.3) In any model for neutral geometry either the Euclidean Parallel Postulate or the Hyperbolic Parallel Postulate will hold.

The Universal Hyperbolic Theorem

Just as existence of triangles with defect 0◦ is all or nothing in neutral geometry, so two are unique parallels. That is, as the next theorem asserts, uniqueness of a parallel for one line and external point guarantees unique parallels for all lines and all possible external points.

The Universal Hyperbolic Theorem (Theorem 4.9.1)

If there exists one line `0, an external point 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.

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

Just as existence of triangles with defect 0◦ is all or nothing in neutral geometry, so two are unique parallels. That is, as the next theorem asserts, uniqueness of a parallel for one line and external point guarantees unique parallels for all lines and all possible external points.

The Universal Hyperbolic Theorem (Theorem 4.9.1)

If there exists one line `0, an external point 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) Neutral geometry Fall 2020 Proof.

Assume there exists a line `0, an external point P0, and at least two lines that pass through P0 and are parallel to `0. Let ` be a line and P an external point. We claim there exist at least two lines through P that are parallel to `.

Let m be the line through P parallel to m obtained by the Double Perpendicular Construction and let Q be the foot of that perpendicular. Choose a point R on ` that is different for Q and let t be the line through R that is perpendicular to `. Drop a perpendicular from P to t and call the foot of that perpendicular S.

Now PQRS is a Lambert quadrilateral. By our hypothesis, the Euclidean Parallel Postulate fails and so PQRS is not a rectangle (Corollary 4.8.7). Hence, QPS is  ←→ ←→ ∠ not a right angle (ACP) and so PS 6= m. But PS is parallel to ` (AIAT).

The Universal Hyperbolic Theorem

The Universal Hyperbolic Theorem (Theorem 4.9.1)

If there exists one line `0, an external point 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 `.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Assume there exists a line `0, an external point P0, and at least two lines that pass through P0 and are parallel to `0. Let ` be a line and P an external point. We claim there exist at least two lines through P that are parallel to `.

Let m be the line through P parallel to m obtained by the Double Perpendicular Construction and let Q be the foot of that perpendicular. Choose a point R on ` that is different for Q and let t be the line through R that is perpendicular to `. Drop a perpendicular from P to t and call the foot of that perpendicular S.

Now PQRS is a Lambert quadrilateral. By our hypothesis, the Euclidean Parallel Postulate fails and so PQRS is not a rectangle (Corollary 4.8.7). Hence, QPS is  ←→ ←→ ∠ not a right angle (ACP) and so PS 6= m. But PS is parallel to ` (AIAT).

The Universal Hyperbolic Theorem

The Universal Hyperbolic Theorem (Theorem 4.9.1)

If there exists one line `0, an external point 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 `.

Proof.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Let ` be a line and P an external point. We claim there exist at least two lines through P that are parallel to `.

Let m be the line through P parallel to m obtained by the Double Perpendicular Construction and let Q be the foot of that perpendicular. Choose a point R on ` that is different for Q and let t be the line through R that is perpendicular to `. Drop a perpendicular from P to t and call the foot of that perpendicular S.

Now PQRS is a Lambert quadrilateral. By our hypothesis, the Euclidean Parallel Postulate fails and so PQRS is not a rectangle (Corollary 4.8.7). Hence, QPS is  ←→ ←→ ∠ not a right angle (ACP) and so PS 6= m. But PS is parallel to ` (AIAT).

The Universal Hyperbolic Theorem

The Universal Hyperbolic Theorem (Theorem 4.9.1)

If there exists one line `0, an external point 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 `.

Proof.

Assume there exists a line `0, an external point P0, and at least two lines that pass through P0 and are parallel to `0.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Let m be the line through P parallel to m obtained by the Double Perpendicular Construction and let Q be the foot of that perpendicular. Choose a point R on ` that is different for Q and let t be the line through R that is perpendicular to `. Drop a perpendicular from P to t and call the foot of that perpendicular S.

Now PQRS is a Lambert quadrilateral. By our hypothesis, the Euclidean Parallel Postulate fails and so PQRS is not a rectangle (Corollary 4.8.7). Hence, QPS is  ←→ ←→ ∠ not a right angle (ACP) and so PS 6= m. But PS is parallel to ` (AIAT).

The Universal Hyperbolic Theorem

The Universal Hyperbolic Theorem (Theorem 4.9.1)

If there exists one line `0, an external point 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 `.

Proof.

Assume there exists a line `0, an external point P0, and at least two lines that pass through P0 and are parallel to `0. Let ` be a line and P an external point. We claim there exist at least two lines through P that are parallel to `.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Choose a point R on ` that is different for Q and let t be the line through R that is perpendicular to `. Drop a perpendicular from P to t and call the foot of that perpendicular S.

Now PQRS is a Lambert quadrilateral. By our hypothesis, the Euclidean Parallel Postulate fails and so PQRS is not a rectangle (Corollary 4.8.7). Hence, QPS is  ←→ ←→ ∠ not a right angle (ACP) and so PS 6= m. But PS is parallel to ` (AIAT).

The Universal Hyperbolic Theorem

The Universal Hyperbolic Theorem (Theorem 4.9.1)

If there exists one line `0, an external point 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 `.

Proof.

Assume there exists a line `0, an external point P0, and at least two lines that pass through P0 and are parallel to `0. Let ` be a line and P an external point. We claim there exist at least two lines through P that are parallel to `.

Let m be the line through P parallel to m obtained by the Double Perpendicular Construction and let Q be the foot of that perpendicular.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Drop a perpendicular from P to t and call the foot of that perpendicular S.

Now PQRS is a Lambert quadrilateral. By our hypothesis, the Euclidean Parallel Postulate fails and so PQRS is not a rectangle (Corollary 4.8.7). Hence, QPS is  ←→ ←→ ∠ not a right angle (ACP) and so PS 6= m. But PS is parallel to ` (AIAT).

The Universal Hyperbolic Theorem

The Universal Hyperbolic Theorem (Theorem 4.9.1)

If there exists one line `0, an external point 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 `.

Proof.

Assume there exists a line `0, an external point P0, and at least two lines that pass through P0 and are parallel to `0. Let ` be a line and P an external point. We claim there exist at least two lines through P that are parallel to `.

Let m be the line through P parallel to m obtained by the Double Perpendicular Construction and let Q be the foot of that perpendicular. Choose a point R on ` that is different for Q and let t be the line through R that is perpendicular to `.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Now PQRS is a Lambert quadrilateral. By our hypothesis, the Euclidean Parallel Postulate fails and so PQRS is not a rectangle (Corollary 4.8.7). Hence, QPS is  ←→ ←→ ∠ not a right angle (ACP) and so PS 6= m. But PS is parallel to ` (AIAT).

The Universal Hyperbolic Theorem

The Universal Hyperbolic Theorem (Theorem 4.9.1)

If there exists one line `0, an external point 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 `.

Proof.

Assume there exists a line `0, an external point P0, and at least two lines that pass through P0 and are parallel to `0. Let ` be a line and P an external point. We claim there exist at least two lines through P that are parallel to `.

Let m be the line through P parallel to m obtained by the Double Perpendicular Construction and let Q be the foot of that perpendicular. Choose a point R on ` that is different for Q and let t be the line through R that is perpendicular to `. Drop a perpendicular from P to t and call the foot of that perpendicular S.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 By our hypothesis, the Euclidean Parallel Postulate fails and so PQRS is not a rectangle (Corollary 4.8.7). Hence, QPS is  ←→ ←→ ∠ not a right angle (ACP) and so PS 6= m. But PS is parallel to ` (AIAT).

The Universal Hyperbolic Theorem

The Universal Hyperbolic Theorem (Theorem 4.9.1)

If there exists one line `0, an external point 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 `.

Proof.

Assume there exists a line `0, an external point P0, and at least two lines that pass through P0 and are parallel to `0. Let ` be a line and P an external point. We claim there exist at least two lines through P that are parallel to `.

Let m be the line through P parallel to m obtained by the Double Perpendicular Construction and let Q be the foot of that perpendicular. Choose a point R on ` that is different for Q and let t be the line through R that is perpendicular to `. Drop a perpendicular from P to t and call the foot of that perpendicular S.

Now PQRS is a Lambert quadrilateral.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Hence, QPS is ←→ ←→ ∠ not a right angle (ACP) and so PS 6= m. But PS is parallel to ` (AIAT).

The Universal Hyperbolic Theorem

The Universal Hyperbolic Theorem (Theorem 4.9.1)

If there exists one line `0, an external point 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 `.

Proof.

Assume there exists a line `0, an external point P0, and at least two lines that pass through P0 and are parallel to `0. Let ` be a line and P an external point. We claim there exist at least two lines through P that are parallel to `.

Let m be the line through P parallel to m obtained by the Double Perpendicular Construction and let Q be the foot of that perpendicular. Choose a point R on ` that is different for Q and let t be the line through R that is perpendicular to `. Drop a perpendicular from P to t and call the foot of that perpendicular S.

Now PQRS is a Lambert quadrilateral. By our hypothesis, the Euclidean Parallel Postulate fails and so PQRS is not a rectangle (Corollary 4.8.7).

MTH 411/511 (Geometry) Neutral geometry Fall 2020 ←→ But PS is parallel to ` (AIAT).

The Universal Hyperbolic Theorem

The Universal Hyperbolic Theorem (Theorem 4.9.1)

If there exists one line `0, an external point 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 `.

Proof.

Assume there exists a line `0, an external point P0, and at least two lines that pass through P0 and are parallel to `0. Let ` be a line and P an external point. We claim there exist at least two lines through P that are parallel to `.

Let m be the line through P parallel to m obtained by the Double Perpendicular Construction and let Q be the foot of that perpendicular. Choose a point R on ` that is different for Q and let t be the line through R that is perpendicular to `. Drop a perpendicular from P to t and call the foot of that perpendicular S.

Now PQRS is a Lambert quadrilateral. By our hypothesis, the Euclidean Parallel Postulate fails and so PQRS is not a rectangle (Corollary 4.8.7). Hence, QPS is  ←→ ∠ not a right angle (ACP) and so PS 6= m.

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

The Universal Hyperbolic Theorem (Theorem 4.9.1)

If there exists one line `0, an external point 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 `.

Proof.

Assume there exists a line `0, an external point P0, and at least two lines that pass through P0 and are parallel to `0. Let ` be a line and P an external point. We claim there exist at least two lines through P that are parallel to `.

Let m be the line through P parallel to m obtained by the Double Perpendicular Construction and let Q be the foot of that perpendicular. Choose a point R on ` that is different for Q and let t be the line through R that is perpendicular to `. Drop a perpendicular from P to t and call the foot of that perpendicular S.

Now PQRS is a Lambert quadrilateral. By our hypothesis, the Euclidean Parallel Postulate fails and so PQRS is not a rectangle (Corollary 4.8.7). Hence, QPS is  ←→ ←→ ∠ not a right angle (ACP) and so PS 6= m. But PS is parallel to ` (AIAT).

MTH 411/511 (Geometry) Neutral geometry Fall 2020 Before next class: Read Sections 5.1 and 5.2.

In the next lecture we will: • Begin our study of Euclidean geometry. • State and prove the Parallel Projection Theorem.

Next time

MTH 411/511 (Geometry) Neutral geometry Fall 2020 In the next lecture we will: • Begin our study of Euclidean geometry. • State and prove the Parallel Projection Theorem.

Next time

Before next class: Read Sections 5.1 and 5.2.

MTH 411/511 (Geometry) Neutral geometry Fall 2020 • Begin our study of Euclidean geometry. • State and prove the Parallel Projection Theorem.

Next time

Before next class: Read Sections 5.1 and 5.2.

In the next lecture we will:

MTH 411/511 (Geometry) Neutral geometry Fall 2020 • State and prove the Parallel Projection Theorem.

Next time

Before next class: Read Sections 5.1 and 5.2.

In the next lecture we will: • Begin our study of Euclidean geometry.

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

Before next class: Read Sections 5.1 and 5.2.

In the next lecture we will: • Begin our study of Euclidean geometry. • State and prove the Parallel Projection Theorem.

MTH 411/511 (Geometry) Neutral geometry Fall 2020