Chapter 5: Euclidean geometry §5.1 Basic Theorems of Euclidean Geometry §5.2 The Parallel Projection Theorem

MTH 411/511

Foundations of Geometry

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

It’s good to have goals

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

It’s good to have goals

Goals for today:

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

It’s good to have goals

Goals for today: • Begin our study of Euclidean geometry.

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

Goals for today: • Begin our study of Euclidean geometry. • State and prove the Parallel Projection Theorem.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Pythagorean Theorem (Theorem 5.4.1) If 4ABC is a right triangle with right angle at vertex C, then

(AC)2 + (BC)2 = (AB)2.

Euclidean geometry

We now add the Euclidean Parallel Postulate to our list of axioms. Our goal in this chapter will be to develop the theory sufficiently so as to prove the following fundamental results. Fundamental Theorem on Similar Triangles (Theorem 5.3.1) If 4ABC and 4DEF are two triangles such that 4ABC ∼ 4DEF , then AB DE = . AC DF

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

We now add the Euclidean Parallel Postulate to our list of axioms. Our goal in this chapter will be to develop the theory sufficiently so as to prove the following fundamental results. Fundamental Theorem on Similar Triangles (Theorem 5.3.1) If 4ABC and 4DEF are two triangles such that 4ABC ∼ 4DEF , then AB DE = . AC DF

Pythagorean Theorem (Theorem 5.4.1) If 4ABC is a right triangle with right angle at vertex C, then

(AC)2 + (BC)2 = (AB)2.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Converse to the AIAT (Theorem 5.1.1) If two parallel lines are cut by a transversal, then both pairs of alternate interior angles are congruent.

Euclid’s Postulate V (Theorem 5.1.2) If ` and `0 are two lines cut by a transversal t in such a way that the sum of the measures of the two interior angles on one side of t is less that 180◦, then ` and `0 intersect on that side of t.

Angle Sum Theorem (5.1.3) If 4ABC is a triangle, then σ(4ABC) = 180◦.

Wallis’ Postulate If 4ABC is a triangle and DE is a segment, then there exists a point F such that 4ABC ∼ 4DEF

Euclidean geometry

Since we are assuming the Euclidean Parallel Postulate, the following statements which we proved equivalent are now all theorems in Euclidean Geometry.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Euclid’s Postulate V (Theorem 5.1.2) If ` and `0 are two lines cut by a transversal t in such a way that the sum of the measures of the two interior angles on one side of t is less that 180◦, then ` and `0 intersect on that side of t.

Angle Sum Theorem (5.1.3) If 4ABC is a triangle, then σ(4ABC) = 180◦.

Wallis’ Postulate If 4ABC is a triangle and DE is a segment, then there exists a point F such that 4ABC ∼ 4DEF

Euclidean geometry

Since we are assuming the Euclidean Parallel Postulate, the following statements which we proved equivalent are now all theorems in Euclidean Geometry. Converse to the AIAT (Theorem 5.1.1) If two parallel lines are cut by a transversal, then both pairs of alternate interior angles are congruent.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Angle Sum Theorem (5.1.3) If 4ABC is a triangle, then σ(4ABC) = 180◦.

Wallis’ Postulate If 4ABC is a triangle and DE is a segment, then there exists a point F such that 4ABC ∼ 4DEF

Euclidean geometry

Since we are assuming the Euclidean Parallel Postulate, the following statements which we proved equivalent are now all theorems in Euclidean Geometry. Converse to the AIAT (Theorem 5.1.1) If two parallel lines are cut by a transversal, then both pairs of alternate interior angles are congruent.

Euclid’s Postulate V (Theorem 5.1.2) If ` and `0 are two lines cut by a transversal t in such a way that the sum of the measures of the two interior angles on one side of t is less that 180◦, then ` and `0 intersect on that side of t.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Wallis’ Postulate If 4ABC is a triangle and DE is a segment, then there exists a point F such that 4ABC ∼ 4DEF

Euclidean geometry

Since we are assuming the Euclidean Parallel Postulate, the following statements which we proved equivalent are now all theorems in Euclidean Geometry. Converse to the AIAT (Theorem 5.1.1) If two parallel lines are cut by a transversal, then both pairs of alternate interior angles are congruent.

Euclid’s Postulate V (Theorem 5.1.2) If ` and `0 are two lines cut by a transversal t in such a way that the sum of the measures of the two interior angles on one side of t is less that 180◦, then ` and `0 intersect on that side of t.

Angle Sum Theorem (5.1.3) If 4ABC is a triangle, then σ(4ABC) = 180◦.

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

Since we are assuming the Euclidean Parallel Postulate, the following statements which we proved equivalent are now all theorems in Euclidean Geometry. Converse to the AIAT (Theorem 5.1.1) If two parallel lines are cut by a transversal, then both pairs of alternate interior angles are congruent.

Euclid’s Postulate V (Theorem 5.1.2) If ` and `0 are two lines cut by a transversal t in such a way that the sum of the measures of the two interior angles on one side of t is less that 180◦, then ` and `0 intersect on that side of t.

Angle Sum Theorem (5.1.3) If 4ABC is a triangle, then σ(4ABC) = 180◦.

Wallis’ Postulate If 4ABC is a triangle and DE is a segment, then there exists a point F such that 4ABC ∼ 4DEF

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Proclus’s Axiom (Theorem 5.1.5) If ` and `0 are parallel lines and t 6= ` is a line such that t intersects `, then t also intersects `0.

(Theorem 5.1.6) If ` and `0 are parallel lines and t is a transversal such that t ⊥ `, then t ⊥ `0.

(Theorem 5.1.7) If `, m, n and k are lines such that k k `, m ⊥ k, and n ⊥ `, then either m = n or m k n.

Transitivity of Parallelism (Theorem 5.1.8) If ` is parallel to m and m is parallel to n, then either ` = n or ` k n.

Euclidean geometry

Since we are assuming the Euclidean Parallel Postulate, the following statements which we proved equivalent are now all theorems in Euclidean Geometry.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 (Theorem 5.1.6) If ` and `0 are parallel lines and t is a transversal such that t ⊥ `, then t ⊥ `0.

(Theorem 5.1.7) If `, m, n and k are lines such that k k `, m ⊥ k, and n ⊥ `, then either m = n or m k n.

Transitivity of Parallelism (Theorem 5.1.8) If ` is parallel to m and m is parallel to n, then either ` = n or ` k n.

Euclidean geometry

Since we are assuming the Euclidean Parallel Postulate, the following statements which we proved equivalent are now all theorems in Euclidean Geometry.

Proclus’s Axiom (Theorem 5.1.5) If ` and `0 are parallel lines and t 6= ` is a line such that t intersects `, then t also intersects `0.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 (Theorem 5.1.7) If `, m, n and k are lines such that k k `, m ⊥ k, and n ⊥ `, then either m = n or m k n.

Transitivity of Parallelism (Theorem 5.1.8) If ` is parallel to m and m is parallel to n, then either ` = n or ` k n.

Euclidean geometry

Since we are assuming the Euclidean Parallel Postulate, the following statements which we proved equivalent are now all theorems in Euclidean Geometry.

Proclus’s Axiom (Theorem 5.1.5) If ` and `0 are parallel lines and t 6= ` is a line such that t intersects `, then t also intersects `0.

(Theorem 5.1.6) If ` and `0 are parallel lines and t is a transversal such that t ⊥ `, then t ⊥ `0.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Transitivity of Parallelism (Theorem 5.1.8) If ` is parallel to m and m is parallel to n, then either ` = n or ` k n.

Euclidean geometry

Since we are assuming the Euclidean Parallel Postulate, the following statements which we proved equivalent are now all theorems in Euclidean Geometry.

Proclus’s Axiom (Theorem 5.1.5) If ` and `0 are parallel lines and t 6= ` is a line such that t intersects `, then t also intersects `0.

(Theorem 5.1.6) If ` and `0 are parallel lines and t is a transversal such that t ⊥ `, then t ⊥ `0.

(Theorem 5.1.7) If `, m, n and k are lines such that k k `, m ⊥ k, and n ⊥ `, then either m = n or m k n.

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

Since we are assuming the Euclidean Parallel Postulate, the following statements which we proved equivalent are now all theorems in Euclidean Geometry.

Proclus’s Axiom (Theorem 5.1.5) If ` and `0 are parallel lines and t 6= ` is a line such that t intersects `, then t also intersects `0.

(Theorem 5.1.6) If ` and `0 are parallel lines and t is a transversal such that t ⊥ `, then t ⊥ `0.

(Theorem 5.1.7) If `, m, n and k are lines such that k k `, m ⊥ k, and n ⊥ `, then either m = n or m k n.

Transitivity of Parallelism (Theorem 5.1.8) If ` is parallel to m and m is parallel to n, then either ` = n or ` k n.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Clairaut’s Axiom (Theorem 5.1.9) There exists a rectangle.

Proofs of the next results are left as an exercise. Properties of Euclidean Parallelograms (Theorem 5.1.10) If ABCD is a parallelogram, then 1. the diagonals divide the quadrilateral into congruent triangles (i.e., 4ABC =∼ 4CDA and 4ABD =∼ 4CDB), 2. the opposite sides are congruent (i.e., AB =∼ CD and BC =∼ AD), ∼ ∼ 3. the opposite angles are congruent (I.e. ∠DAB = ∠BCD and ∠ABC = ∠CDA), 4. the diagonals bisect each other (i.e., AC and BD intersect in a point E that is the midpoint of each).

Euclidean geometry

Since we are assuming the Euclidean Parallel Postulate, the following statements which we proved equivalent are now all theorems in Euclidean Geometry.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Proofs of the next results are left as an exercise. Properties of Euclidean Parallelograms (Theorem 5.1.10) If ABCD is a parallelogram, then 1. the diagonals divide the quadrilateral into congruent triangles (i.e., 4ABC =∼ 4CDA and 4ABD =∼ 4CDB), 2. the opposite sides are congruent (i.e., AB =∼ CD and BC =∼ AD), ∼ ∼ 3. the opposite angles are congruent (I.e. ∠DAB = ∠BCD and ∠ABC = ∠CDA), 4. the diagonals bisect each other (i.e., AC and BD intersect in a point E that is the midpoint of each).

Euclidean geometry

Since we are assuming the Euclidean Parallel Postulate, the following statements which we proved equivalent are now all theorems in Euclidean Geometry.

Clairaut’s Axiom (Theorem 5.1.9) There exists a rectangle.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Properties of Euclidean Parallelograms (Theorem 5.1.10) If ABCD is a parallelogram, then 1. the diagonals divide the quadrilateral into congruent triangles (i.e., 4ABC =∼ 4CDA and 4ABD =∼ 4CDB), 2. the opposite sides are congruent (i.e., AB =∼ CD and BC =∼ AD), ∼ ∼ 3. the opposite angles are congruent (I.e. ∠DAB = ∠BCD and ∠ABC = ∠CDA), 4. the diagonals bisect each other (i.e., AC and BD intersect in a point E that is the midpoint of each).

Euclidean geometry

Since we are assuming the Euclidean Parallel Postulate, the following statements which we proved equivalent are now all theorems in Euclidean Geometry.

Clairaut’s Axiom (Theorem 5.1.9) There exists a rectangle.

Proofs of the next results are left as an exercise.

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

Since we are assuming the Euclidean Parallel Postulate, the following statements which we proved equivalent are now all theorems in Euclidean Geometry.

Clairaut’s Axiom (Theorem 5.1.9) There exists a rectangle.

Proofs of the next results are left as an exercise. Properties of Euclidean Parallelograms (Theorem 5.1.10) If ABCD is a parallelogram, then 1. the diagonals divide the quadrilateral into congruent triangles (i.e., 4ABC =∼ 4CDA and 4ABD =∼ 4CDB), 2. the opposite sides are congruent (i.e., AB =∼ CD and BC =∼ AD), ∼ ∼ 3. the opposite angles are congruent (I.e. ∠DAB = ∠BCD and ∠ABC = ∠CDA), 4. the diagonals bisect each other (i.e., AC and BD intersect in a point E that is the midpoint of each).

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 First we need a lemma.

The Parallel Projection Theorem

The Parallel Projection Theorem (Theorem 5.2.1) Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C, then AB A0B0 = . AC A0C 0

A A0 `

0 B B m

0 C C n t t0

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 The Parallel Projection Theorem

The Parallel Projection Theorem (Theorem 5.2.1) Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C, then AB A0B0 = . AC A0C 0

A A0 `

0 B B m

0 C C n t t0

First we need a lemma.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Proof. Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C and AB =∼ BC. We claim A0B0 =∼ B0C 0.

If A0 6= A, let t00 be the line through A0 such that t00 k t (EPP). if A0 = A, then let t00 = t. Similarly, if B0 6= B, let t00 be the line through B0 that is parallel to t; otherwise let t000 = t. Define B00 to be the point at which t00 crosses m and C 000 to be the point at which t000 crosses n. Note that all of these points exist by Proclus’s Axiom.

The Parallel Projection Theorem

(Lemma 5.2.2) Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C. If AB =∼ BC, then A0B0 =∼ B0C 0.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 If A0 6= A, let t00 be the line through A0 such that t00 k t (EPP). if A0 = A, then let t00 = t. Similarly, if B0 6= B, let t00 be the line through B0 that is parallel to t; otherwise let t000 = t. Define B00 to be the point at which t00 crosses m and C 000 to be the point at which t000 crosses n. Note that all of these points exist by Proclus’s Axiom.

The Parallel Projection Theorem

(Lemma 5.2.2) Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C. If AB =∼ BC, then A0B0 =∼ B0C 0.

Proof. Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C and AB =∼ BC. We claim A0B0 =∼ B0C 0.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 if A0 = A, then let t00 = t. Similarly, if B0 6= B, let t00 be the line through B0 that is parallel to t; otherwise let t000 = t. Define B00 to be the point at which t00 crosses m and C 000 to be the point at which t000 crosses n. Note that all of these points exist by Proclus’s Axiom.

The Parallel Projection Theorem

(Lemma 5.2.2) Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C. If AB =∼ BC, then A0B0 =∼ B0C 0.

Proof. Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C and AB =∼ BC. We claim A0B0 =∼ B0C 0.

If A0 6= A, let t00 be the line through A0 such that t00 k t (EPP).

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Define B00 to be the point at which t00 crosses m and C 000 to be the point at which t000 crosses n. Note that all of these points exist by Proclus’s Axiom.

The Parallel Projection Theorem

(Lemma 5.2.2) Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C. If AB =∼ BC, then A0B0 =∼ B0C 0.

Proof. Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C and AB =∼ BC. We claim A0B0 =∼ B0C 0.

If A0 6= A, let t00 be the line through A0 such that t00 k t (EPP). if A0 = A, then let t00 = t. Similarly, if B0 6= B, let t00 be the line through B0 that is parallel to t; otherwise let t000 = t.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 The Parallel Projection Theorem

(Lemma 5.2.2) Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C. If AB =∼ BC, then A0B0 =∼ B0C 0.

Proof. Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C and AB =∼ BC. We claim A0B0 =∼ B0C 0.

If A0 6= A, let t00 be the line through A0 such that t00 k t (EPP). if A0 = A, then let t00 = t. Similarly, if B0 6= B, let t00 be the line through B0 that is parallel to t; otherwise let t000 = t. Define B00 to be the point at which t00 crosses m and C 000 to be the point at which t000 crosses n. Note that all of these points exist by Proclus’s Axiom.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Proof. If A0 = A, then B00 = B and AB = A0B00. Similarly, if B0 = B, then C 000 = C and BC = B0C 000. If A0 6= A and B0 6= B, then A0B00 =∼ AB and B0C 000 =∼ BC (Properties of Euclidean Parallelograms). Since AB =∼ BC (hypothesis), then A0B00 =∼ B0C 000 (in all cases).

If B00 = B0, then C 000 = C and the proof is complete. Otherwise, t00 k t000 or t00 = t000 00 0 0 ∼ 000 0 0 0 00 0 ∼ 0 000 0 (Transitivity of Parallelism). Hence, ∠B A B = ∠C B C and ∠A B B = ∠B C C (Converse to Corresponding Angles Theorem). Thus, 4B00A0B0 =∼ 4C 000B0C 0 (ASA). It follows that A0B0 =∼ B0C 0 (definition of congruent triangles).

The Parallel Projection Theorem

(Lemma 5.2.2) Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C. If AB =∼ BC, then A0B0 =∼ B0C 0.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Similarly, if B0 = B, then C 000 = C and BC = B0C 000. If A0 6= A and B0 6= B, then A0B00 =∼ AB and B0C 000 =∼ BC (Properties of Euclidean Parallelograms). Since AB =∼ BC (hypothesis), then A0B00 =∼ B0C 000 (in all cases).

If B00 = B0, then C 000 = C and the proof is complete. Otherwise, t00 k t000 or t00 = t000 00 0 0 ∼ 000 0 0 0 00 0 ∼ 0 000 0 (Transitivity of Parallelism). Hence, ∠B A B = ∠C B C and ∠A B B = ∠B C C (Converse to Corresponding Angles Theorem). Thus, 4B00A0B0 =∼ 4C 000B0C 0 (ASA). It follows that A0B0 =∼ B0C 0 (definition of congruent triangles).

The Parallel Projection Theorem

(Lemma 5.2.2) Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C. If AB =∼ BC, then A0B0 =∼ B0C 0.

Proof. If A0 = A, then B00 = B and AB = A0B00.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 If A0 6= A and B0 6= B, then A0B00 =∼ AB and B0C 000 =∼ BC (Properties of Euclidean Parallelograms). Since AB =∼ BC (hypothesis), then A0B00 =∼ B0C 000 (in all cases).

If B00 = B0, then C 000 = C and the proof is complete. Otherwise, t00 k t000 or t00 = t000 00 0 0 ∼ 000 0 0 0 00 0 ∼ 0 000 0 (Transitivity of Parallelism). Hence, ∠B A B = ∠C B C and ∠A B B = ∠B C C (Converse to Corresponding Angles Theorem). Thus, 4B00A0B0 =∼ 4C 000B0C 0 (ASA). It follows that A0B0 =∼ B0C 0 (definition of congruent triangles).

The Parallel Projection Theorem

(Lemma 5.2.2) Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C. If AB =∼ BC, then A0B0 =∼ B0C 0.

Proof. If A0 = A, then B00 = B and AB = A0B00. Similarly, if B0 = B, then C 000 = C and BC = B0C 000.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Since AB =∼ BC (hypothesis), then A0B00 =∼ B0C 000 (in all cases).

If B00 = B0, then C 000 = C and the proof is complete. Otherwise, t00 k t000 or t00 = t000 00 0 0 ∼ 000 0 0 0 00 0 ∼ 0 000 0 (Transitivity of Parallelism). Hence, ∠B A B = ∠C B C and ∠A B B = ∠B C C (Converse to Corresponding Angles Theorem). Thus, 4B00A0B0 =∼ 4C 000B0C 0 (ASA). It follows that A0B0 =∼ B0C 0 (definition of congruent triangles).

The Parallel Projection Theorem

(Lemma 5.2.2) Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C. If AB =∼ BC, then A0B0 =∼ B0C 0.

Proof. If A0 = A, then B00 = B and AB = A0B00. Similarly, if B0 = B, then C 000 = C and BC = B0C 000. If A0 6= A and B0 6= B, then A0B00 =∼ AB and B0C 000 =∼ BC (Properties of Euclidean Parallelograms).

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 If B00 = B0, then C 000 = C and the proof is complete. Otherwise, t00 k t000 or t00 = t000 00 0 0 ∼ 000 0 0 0 00 0 ∼ 0 000 0 (Transitivity of Parallelism). Hence, ∠B A B = ∠C B C and ∠A B B = ∠B C C (Converse to Corresponding Angles Theorem). Thus, 4B00A0B0 =∼ 4C 000B0C 0 (ASA). It follows that A0B0 =∼ B0C 0 (definition of congruent triangles).

The Parallel Projection Theorem

(Lemma 5.2.2) Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C. If AB =∼ BC, then A0B0 =∼ B0C 0.

Proof. If A0 = A, then B00 = B and AB = A0B00. Similarly, if B0 = B, then C 000 = C and BC = B0C 000. If A0 6= A and B0 6= B, then A0B00 =∼ AB and B0C 000 =∼ BC (Properties of Euclidean Parallelograms). Since AB =∼ BC (hypothesis), then A0B00 =∼ B0C 000 (in all cases).

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Otherwise, t00 k t000 or t00 = t000 00 0 0 ∼ 000 0 0 0 00 0 ∼ 0 000 0 (Transitivity of Parallelism). Hence, ∠B A B = ∠C B C and ∠A B B = ∠B C C (Converse to Corresponding Angles Theorem). Thus, 4B00A0B0 =∼ 4C 000B0C 0 (ASA). It follows that A0B0 =∼ B0C 0 (definition of congruent triangles).

The Parallel Projection Theorem

(Lemma 5.2.2) Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C. If AB =∼ BC, then A0B0 =∼ B0C 0.

Proof. If A0 = A, then B00 = B and AB = A0B00. Similarly, if B0 = B, then C 000 = C and BC = B0C 000. If A0 6= A and B0 6= B, then A0B00 =∼ AB and B0C 000 =∼ BC (Properties of Euclidean Parallelograms). Since AB =∼ BC (hypothesis), then A0B00 =∼ B0C 000 (in all cases).

If B00 = B0, then C 000 = C and the proof is complete.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 00 0 0 ∼ 000 0 0 0 00 0 ∼ 0 000 0 Hence, ∠B A B = ∠C B C and ∠A B B = ∠B C C (Converse to Corresponding Angles Theorem). Thus, 4B00A0B0 =∼ 4C 000B0C 0 (ASA). It follows that A0B0 =∼ B0C 0 (definition of congruent triangles).

The Parallel Projection Theorem

(Lemma 5.2.2) Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C. If AB =∼ BC, then A0B0 =∼ B0C 0.

Proof. If A0 = A, then B00 = B and AB = A0B00. Similarly, if B0 = B, then C 000 = C and BC = B0C 000. If A0 6= A and B0 6= B, then A0B00 =∼ AB and B0C 000 =∼ BC (Properties of Euclidean Parallelograms). Since AB =∼ BC (hypothesis), then A0B00 =∼ B0C 000 (in all cases).

If B00 = B0, then C 000 = C and the proof is complete. Otherwise, t00 k t000 or t00 = t000 (Transitivity of Parallelism).

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Thus, 4B00A0B0 =∼ 4C 000B0C 0 (ASA). It follows that A0B0 =∼ B0C 0 (definition of congruent triangles).

The Parallel Projection Theorem

(Lemma 5.2.2) Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C. If AB =∼ BC, then A0B0 =∼ B0C 0.

Proof. If A0 = A, then B00 = B and AB = A0B00. Similarly, if B0 = B, then C 000 = C and BC = B0C 000. If A0 6= A and B0 6= B, then A0B00 =∼ AB and B0C 000 =∼ BC (Properties of Euclidean Parallelograms). Since AB =∼ BC (hypothesis), then A0B00 =∼ B0C 000 (in all cases).

If B00 = B0, then C 000 = C and the proof is complete. Otherwise, t00 k t000 or t00 = t000 00 0 0 ∼ 000 0 0 0 00 0 ∼ 0 000 0 (Transitivity of Parallelism). Hence, ∠B A B = ∠C B C and ∠A B B = ∠B C C (Converse to Corresponding Angles Theorem).

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 It follows that A0B0 =∼ B0C 0 (definition of congruent triangles).

The Parallel Projection Theorem

(Lemma 5.2.2) Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C. If AB =∼ BC, then A0B0 =∼ B0C 0.

Proof. If A0 = A, then B00 = B and AB = A0B00. Similarly, if B0 = B, then C 000 = C and BC = B0C 000. If A0 6= A and B0 6= B, then A0B00 =∼ AB and B0C 000 =∼ BC (Properties of Euclidean Parallelograms). Since AB =∼ BC (hypothesis), then A0B00 =∼ B0C 000 (in all cases).

If B00 = B0, then C 000 = C and the proof is complete. Otherwise, t00 k t000 or t00 = t000 00 0 0 ∼ 000 0 0 0 00 0 ∼ 0 000 0 (Transitivity of Parallelism). Hence, ∠B A B = ∠C B C and ∠A B B = ∠B C C (Converse to Corresponding Angles Theorem). Thus, 4B00A0B0 =∼ 4C 000B0C 0 (ASA).

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 The Parallel Projection Theorem

(Lemma 5.2.2) Let `, m and n be distinct parallel lines. Let t be transversal that cuts these lines at points A, B, and C, respectively, and let t0 be a transversal that cuts these lines at points A0, B0, and C 0, respectively. Assume A ∗ B ∗ C. If AB =∼ BC, then A0B0 =∼ B0C 0.

Proof. If A0 = A, then B00 = B and AB = A0B00. Similarly, if B0 = B, then C 000 = C and BC = B0C 000. If A0 6= A and B0 6= B, then A0B00 =∼ AB and B0C 000 =∼ BC (Properties of Euclidean Parallelograms). Since AB =∼ BC (hypothesis), then A0B00 =∼ B0C 000 (in all cases).

If B00 = B0, then C 000 = C and the proof is complete. Otherwise, t00 k t000 or t00 = t000 00 0 0 ∼ 000 0 0 0 00 0 ∼ 0 000 0 (Transitivity of Parallelism). Hence, ∠B A B = ∠C B C and ∠A B B = ∠B C C (Converse to Corresponding Angles Theorem). Thus, 4B00A0B0 =∼ 4C 000B0C 0 (ASA). It follows that A0B0 =∼ B0C 0 (definition of congruent triangles).

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 We will not prove this result here, though it is proved in the appendix of the book and makes use of the Density Theorem.

Density Theorem (Theorem E.3.2) If a and b are real numbers such that a < b, then there exists a rational number x such that a < x < b and there exists an irrational number y such that a < y < b.

The Comparison Theorem

We will also need the following property of real numbers.

Comparison Theorem (Theorem E.3.3) If x and y are any real numbers such that (1) every rational number that is less that x is also less than y, and (2) every rational number that is less than y is also less than x, then x = y.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Density Theorem (Theorem E.3.2) If a and b are real numbers such that a < b, then there exists a rational number x such that a < x < b and there exists an irrational number y such that a < y < b.

The Comparison Theorem

We will also need the following property of real numbers.

Comparison Theorem (Theorem E.3.3) If x and y are any real numbers such that (1) every rational number that is less that x is also less than y, and (2) every rational number that is less than y is also less than x, then x = y.

We will not prove this result here, though it is proved in the appendix of the book and makes use of the Density Theorem.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 The Comparison Theorem

We will also need the following property of real numbers.

Comparison Theorem (Theorem E.3.3) If x and y are any real numbers such that (1) every rational number that is less that x is also less than y, and (2) every rational number that is less than y is also less than x, then x = y.

We will not prove this result here, though it is proved in the appendix of the book and makes use of the Density Theorem.

Density Theorem (Theorem E.3.2) If a and b are real numbers such that a < b, then there exists a rational number x such that a < x < b and there exists an irrational number y such that a < y < b.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Choose points A0, A1,..., Aq on t such that A0 = A, Aq = C, and for each i, Ai Ai+1 = AC/q (Ruler Postulate). Then Ap = B and for each 0 0 i, 1 ≤ i ≤ q, there exists a line `i such that Ai lies on `i and `i k ` (DPC). Let A0 = A 0 0 and let Ai , i ≥ 1, be the point at which `i crosses t (Proclus’s Axiom). Then 0 0 0 0 Ai Ai+1 = A C /q for each i (Lemma 5.2.2). Since `p = m and `q = n, we must have 0 0 0 0 Ap = B and Aq = C . It follows that

0 0 0 0 0 0 A B A0Ap (A C /q)p p (AC/q)p A0Ap AB 0 0 = 0 0 = 0 0 = = = = . A C A0Aq (A C /q)q q (AC/q)q A0Aq AC This completes the proof in the case AB/AC is rational.

Parallel Projection Theorem

Proof of the Parallel Projection Theorem. First consider the case that AB/AC is a rational number. That is, AB/AC = p/q for positive integers p and q.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Then Ap = B and for each 0 0 i, 1 ≤ i ≤ q, there exists a line `i such that Ai lies on `i and `i k ` (DPC). Let A0 = A 0 0 and let Ai , i ≥ 1, be the point at which `i crosses t (Proclus’s Axiom). Then 0 0 0 0 Ai Ai+1 = A C /q for each i (Lemma 5.2.2). Since `p = m and `q = n, we must have 0 0 0 0 Ap = B and Aq = C . It follows that

0 0 0 0 0 0 A B A0Ap (A C /q)p p (AC/q)p A0Ap AB 0 0 = 0 0 = 0 0 = = = = . A C A0Aq (A C /q)q q (AC/q)q A0Aq AC This completes the proof in the case AB/AC is rational.

Parallel Projection Theorem

Proof of the Parallel Projection Theorem. First consider the case that AB/AC is a rational number. That is, AB/AC = p/q for positive integers p and q. Choose points A0, A1,..., Aq on t such that A0 = A, Aq = C, and for each i, Ai Ai+1 = AC/q (Ruler Postulate).

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 0 0 Let A0 = A 0 0 and let Ai , i ≥ 1, be the point at which `i crosses t (Proclus’s Axiom). Then 0 0 0 0 Ai Ai+1 = A C /q for each i (Lemma 5.2.2). Since `p = m and `q = n, we must have 0 0 0 0 Ap = B and Aq = C . It follows that

0 0 0 0 0 0 A B A0Ap (A C /q)p p (AC/q)p A0Ap AB 0 0 = 0 0 = 0 0 = = = = . A C A0Aq (A C /q)q q (AC/q)q A0Aq AC This completes the proof in the case AB/AC is rational.

Parallel Projection Theorem

Proof of the Parallel Projection Theorem. First consider the case that AB/AC is a rational number. That is, AB/AC = p/q for positive integers p and q. Choose points A0, A1,..., Aq on t such that A0 = A, Aq = C, and for each i, Ai Ai+1 = AC/q (Ruler Postulate). Then Ap = B and for each i, 1 ≤ i ≤ q, there exists a line `i such that Ai lies on `i and `i k ` (DPC).

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Then 0 0 0 0 Ai Ai+1 = A C /q for each i (Lemma 5.2.2). Since `p = m and `q = n, we must have 0 0 0 0 Ap = B and Aq = C . It follows that

0 0 0 0 0 0 A B A0Ap (A C /q)p p (AC/q)p A0Ap AB 0 0 = 0 0 = 0 0 = = = = . A C A0Aq (A C /q)q q (AC/q)q A0Aq AC This completes the proof in the case AB/AC is rational.

Parallel Projection Theorem

Proof of the Parallel Projection Theorem. First consider the case that AB/AC is a rational number. That is, AB/AC = p/q for positive integers p and q. Choose points A0, A1,..., Aq on t such that A0 = A, Aq = C, and for each i, Ai Ai+1 = AC/q (Ruler Postulate). Then Ap = B and for each 0 0 i, 1 ≤ i ≤ q, there exists a line `i such that Ai lies on `i and `i k ` (DPC). Let A0 = A 0 0 and let Ai , i ≥ 1, be the point at which `i crosses t (Proclus’s Axiom).

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Since `p = m and `q = n, we must have 0 0 0 0 Ap = B and Aq = C . It follows that

0 0 0 0 0 0 A B A0Ap (A C /q)p p (AC/q)p A0Ap AB 0 0 = 0 0 = 0 0 = = = = . A C A0Aq (A C /q)q q (AC/q)q A0Aq AC This completes the proof in the case AB/AC is rational.

Parallel Projection Theorem

Proof of the Parallel Projection Theorem. First consider the case that AB/AC is a rational number. That is, AB/AC = p/q for positive integers p and q. Choose points A0, A1,..., Aq on t such that A0 = A, Aq = C, and for each i, Ai Ai+1 = AC/q (Ruler Postulate). Then Ap = B and for each 0 0 i, 1 ≤ i ≤ q, there exists a line `i such that Ai lies on `i and `i k ` (DPC). Let A0 = A 0 0 and let Ai , i ≥ 1, be the point at which `i crosses t (Proclus’s Axiom). Then 0 0 0 0 Ai Ai+1 = A C /q for each i (Lemma 5.2.2).

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 It follows that

0 0 0 0 0 0 A B A0Ap (A C /q)p p (AC/q)p A0Ap AB 0 0 = 0 0 = 0 0 = = = = . A C A0Aq (A C /q)q q (AC/q)q A0Aq AC This completes the proof in the case AB/AC is rational.

Parallel Projection Theorem

Proof of the Parallel Projection Theorem. First consider the case that AB/AC is a rational number. That is, AB/AC = p/q for positive integers p and q. Choose points A0, A1,..., Aq on t such that A0 = A, Aq = C, and for each i, Ai Ai+1 = AC/q (Ruler Postulate). Then Ap = B and for each 0 0 i, 1 ≤ i ≤ q, there exists a line `i such that Ai lies on `i and `i k ` (DPC). Let A0 = A 0 0 and let Ai , i ≥ 1, be the point at which `i crosses t (Proclus’s Axiom). Then 0 0 0 0 Ai Ai+1 = A C /q for each i (Lemma 5.2.2). Since `p = m and `q = n, we must have 0 0 0 0 Ap = B and Aq = C .

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Parallel Projection Theorem

Proof of the Parallel Projection Theorem. First consider the case that AB/AC is a rational number. That is, AB/AC = p/q for positive integers p and q. Choose points A0, A1,..., Aq on t such that A0 = A, Aq = C, and for each i, Ai Ai+1 = AC/q (Ruler Postulate). Then Ap = B and for each 0 0 i, 1 ≤ i ≤ q, there exists a line `i such that Ai lies on `i and `i k ` (DPC). Let A0 = A 0 0 and let Ai , i ≥ 1, be the point at which `i crosses t (Proclus’s Axiom). Then 0 0 0 0 Ai Ai+1 = A C /q for each i (Lemma 5.2.2). Since `p = m and `q = n, we must have 0 0 0 0 Ap = B and Aq = C . It follows that

0 0 0 0 0 0 A B A0Ap (A C /q)p p (AC/q)p A0Ap AB 0 0 = 0 0 = 0 0 = = = = . A C A0Aq (A C /q)q q (AC/q)q A0Aq AC This completes the proof in the case AB/AC is rational.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Let t be a rational number such that 0 < r < x. Choose a point D on t such that AD/AC = r. Let m0 be the line such that D lies on m0 and m0 k m (DPC), and let D0 be the point at which m0 meets t0. (Proclus’s Axiom) Then A0D0/A0C 0 = r (by the previous part). Since `, m, and m0 are parallel, A0 ∗ D0 ∗ B0 and therefore

A0D0 A0B0 r = < = y. A0C 0 A0C 0 A similar arguments shows that if r is rational number such that 0 < r < y, then r < x. Hence, x = y (Comparison Theorem).

Parallel Projection Theorem

Proof of the Parallel Projection Theorem. Now consider the case that AB/AC = x (a possibly irrational number) and A0B0/A0C 0 = y.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Choose a point D on t such that AD/AC = r. Let m0 be the line such that D lies on m0 and m0 k m (DPC), and let D0 be the point at which m0 meets t0. (Proclus’s Axiom) Then A0D0/A0C 0 = r (by the previous part). Since `, m, and m0 are parallel, A0 ∗ D0 ∗ B0 and therefore

A0D0 A0B0 r = < = y. A0C 0 A0C 0 A similar arguments shows that if r is rational number such that 0 < r < y, then r < x. Hence, x = y (Comparison Theorem).

Parallel Projection Theorem

Proof of the Parallel Projection Theorem. Now consider the case that AB/AC = x (a possibly irrational number) and A0B0/A0C 0 = y. Let t be a rational number such that 0 < r < x.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Let m0 be the line such that D lies on m0 and m0 k m (DPC), and let D0 be the point at which m0 meets t0. (Proclus’s Axiom) Then A0D0/A0C 0 = r (by the previous part). Since `, m, and m0 are parallel, A0 ∗ D0 ∗ B0 and therefore

A0D0 A0B0 r = < = y. A0C 0 A0C 0 A similar arguments shows that if r is rational number such that 0 < r < y, then r < x. Hence, x = y (Comparison Theorem).

Parallel Projection Theorem

Proof of the Parallel Projection Theorem. Now consider the case that AB/AC = x (a possibly irrational number) and A0B0/A0C 0 = y. Let t be a rational number such that 0 < r < x. Choose a point D on t such that AD/AC = r.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Then A0D0/A0C 0 = r (by the previous part). Since `, m, and m0 are parallel, A0 ∗ D0 ∗ B0 and therefore

A0D0 A0B0 r = < = y. A0C 0 A0C 0 A similar arguments shows that if r is rational number such that 0 < r < y, then r < x. Hence, x = y (Comparison Theorem).

Parallel Projection Theorem

Proof of the Parallel Projection Theorem. Now consider the case that AB/AC = x (a possibly irrational number) and A0B0/A0C 0 = y. Let t be a rational number such that 0 < r < x. Choose a point D on t such that AD/AC = r. Let m0 be the line such that D lies on m0 and m0 k m (DPC), and let D0 be the point at which m0 meets t0. (Proclus’s Axiom)

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Since `, m, and m0 are parallel, A0 ∗ D0 ∗ B0 and therefore

A0D0 A0B0 r = < = y. A0C 0 A0C 0 A similar arguments shows that if r is rational number such that 0 < r < y, then r < x. Hence, x = y (Comparison Theorem).

Parallel Projection Theorem

Proof of the Parallel Projection Theorem. Now consider the case that AB/AC = x (a possibly irrational number) and A0B0/A0C 0 = y. Let t be a rational number such that 0 < r < x. Choose a point D on t such that AD/AC = r. Let m0 be the line such that D lies on m0 and m0 k m (DPC), and let D0 be the point at which m0 meets t0. (Proclus’s Axiom) Then A0D0/A0C 0 = r (by the previous part).

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Hence, x = y (Comparison Theorem).

Parallel Projection Theorem

Proof of the Parallel Projection Theorem. Now consider the case that AB/AC = x (a possibly irrational number) and A0B0/A0C 0 = y. Let t be a rational number such that 0 < r < x. Choose a point D on t such that AD/AC = r. Let m0 be the line such that D lies on m0 and m0 k m (DPC), and let D0 be the point at which m0 meets t0. (Proclus’s Axiom) Then A0D0/A0C 0 = r (by the previous part). Since `, m, and m0 are parallel, A0 ∗ D0 ∗ B0 and therefore

A0D0 A0B0 r = < = y. A0C 0 A0C 0 A similar arguments shows that if r is rational number such that 0 < r < y, then r < x.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Parallel Projection Theorem

Proof of the Parallel Projection Theorem. Now consider the case that AB/AC = x (a possibly irrational number) and A0B0/A0C 0 = y. Let t be a rational number such that 0 < r < x. Choose a point D on t such that AD/AC = r. Let m0 be the line such that D lies on m0 and m0 k m (DPC), and let D0 be the point at which m0 meets t0. (Proclus’s Axiom) Then A0D0/A0C 0 = r (by the previous part). Since `, m, and m0 are parallel, A0 ∗ D0 ∗ B0 and therefore

A0D0 A0B0 r = < = y. A0C 0 A0C 0 A similar arguments shows that if r is rational number such that 0 < r < y, then r < x. Hence, x = y (Comparison Theorem).

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 Before next class: Read Section 5.3.

In the next lecture we will: • Prove the Fundamental Theorem on Similar Triangles. • Consider its consequences, including the SAS Similarity Criterion. • Use the Fundamental Theorem to prove the Pythagorean Theorem.

Next time

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 In the next lecture we will: • Prove the Fundamental Theorem on Similar Triangles. • Consider its consequences, including the SAS Similarity Criterion. • Use the Fundamental Theorem to prove the Pythagorean Theorem.

Next time

Before next class: Read Section 5.3.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 • Prove the Fundamental Theorem on Similar Triangles. • Consider its consequences, including the SAS Similarity Criterion. • Use the Fundamental Theorem to prove the Pythagorean Theorem.

Next time

Before next class: Read Section 5.3.

In the next lecture we will:

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 • Consider its consequences, including the SAS Similarity Criterion. • Use the Fundamental Theorem to prove the Pythagorean Theorem.

Next time

Before next class: Read Section 5.3.

In the next lecture we will: • Prove the Fundamental Theorem on Similar Triangles.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020 • Use the Fundamental Theorem to prove the Pythagorean Theorem.

Next time

Before next class: Read Section 5.3.

In the next lecture we will: • Prove the Fundamental Theorem on Similar Triangles. • Consider its consequences, including the SAS Similarity Criterion.

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

Before next class: Read Section 5.3.

In the next lecture we will: • Prove the Fundamental Theorem on Similar Triangles. • Consider its consequences, including the SAS Similarity Criterion. • Use the Fundamental Theorem to prove the Pythagorean Theorem.

MTH 411/511 (Geometry) Euclidean geometry Fall 2020