Chapter 4: Neutral §4.1 The Exterior and Existence of Perpendiculars

MTH 411/511

Foundations of Geometry

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

Goals for today: • Review definition of perpendicular and uniqueness of perpendiculars through a on a . • Prove existence and uniqueness of perpendiculars (in general). • Discuss various conditions

MTH 411/511 (Geometry) Euclid’s Elements Fall 2020 Last Time

Last time we proved the following theorem:

Exterior Angle Theorem (Theorem 4.1.2) The measure of an exterior angle for a triangle is strictly greater than the measure of either remote interior angle.

A

D B C E

Today we’ll prove another fundamental result in neutral geometry.

MTH 411/511 (Geometry) Euclid’s Elements Fall 2020 Perpendicular lines

We’ve previously defined perpendicular lines. Since they are critical to our discussion today, we’ll review the definition briefly. Definition 1 Two lines ` and m are perpendicular if there exists a point A that lies on both ` and m and there exist points B ∈ ` and C ∈ m such that ∠BAC is a . Our notation for perpendicular lines is ` ⊥ m.

m C

` P B

MTH 411/511 (Geometry) Euclid’s Elements Fall 2020 Perpendicular lines

(Theorem 3.5.9) If ` is a line and P is a point on `, then there exists exactly one line m such that P lies on m and m ⊥ `.

The proof of the above theorem was essentially just an application of the ACP. Now we prove the same result but without the assumption that P ∈ `. This is called dropping a perpendicular from P to `. The point where m intersects ` is the foot of the perpendicular.

Existence and Uniqueness of Perpendiculars (Theorem 4.1.3) For every line ` and for every point P, there exists a unique line m such that P lies on m and m ⊥ `.

MTH 411/511 (Geometry) Euclid’s Elements Fall 2020 Perpendiculars

Existence and Uniqueness of Perpendiculars (Theorem 4.1.3) For every line ` and for every point P, there exists a unique line m such that P lies on m and m ⊥ `.

Proof. First we prove existence. There exist distinct points Q, Q0 ∈ ` (Ruler Postulate). 0 ∼ 0 There exists a point R, on the opposite side of ` from P, such that ∠Q QP = ∠Q QR −→ ←→ (ACP). Choose a point P0 on QR such that QP =∼ QP0 (PCP). Let m = PP0. We claim that m ⊥ `. −−→ Let F ∈ PP0 ∩ `. There are three cases: F = Q, F 6= Q and F ∈ QQ0, or F 6= Q and −−→ F is on the ray opposite QQ0.

−→ −−→0 0 0 0 Suppose F = Q (PSP). Then QP and QP are opposite rays so ∠Q FP and ∠Q FP form a linear pair (definition of linear pair). Since they are congruent, then they are each right (Linear Pair Theorem). Hence, m ⊥ `.

MTH 411/511 (Geometry) Euclid’s Elements Fall 2020 Perpendiculars

Existence and Uniqueness of Perpendiculars (Theorem 4.1.3) For every line ` and for every point P, there exists a unique line m such that P lies on m and m ⊥ `.

Proof.

−−→0 0 0 0 0 Suppose F 6= Q but F ∈ QQ . Note that ∠PQF = ∠PQQ , ∠P QF = ∠P QQ . By 0 ∼ 0 ∼ 0 our choice of P , ∠PQF = ∠P QF . Hence, 4FQP = FQP (SAS), and so ∼ 0 0 ∠QFP = ∠QFP (definition of congruent ). Since ∠QFP and ∠QFP form a linear pair, then they are right angles by the same as above.

−−→0 0 Finally, suppose F 6= Q and F lies on the ray opposite QQ . Then ∠PQF and ∠PQQ 0 0 0 ∼ 0 are supplements, as are ∠P QF and ∠P QQ . Thus, 4FQP = FQP (SAS) and the proof proceeds as above.

MTH 411/511 (Geometry) Euclid’s Elements Fall 2020 Perpendiculars

Existence and Uniqueness of Perpendiculars (Theorem 4.1.3) For every line ` and for every point P, there exists a unique line m such that P lies on m and m ⊥ `.

Proof. It is left to prove uniqueness. Let m0 be a line such that P lies on m0 and m0 ⊥ `. Let Q0 be the foot of the perpendicular. If Q0 = Q, then m = m0 (Incidence Postulate). −−→ Suppose Q0 6= Q (RAA hypothesis). Let G be a point on QQ0 such that Q ∗ Q0 ∗ G 0 0 (Ruler Postulate). Then ∠PQQ and ∠PQ G are right angles (definition of 0 0 perpendicular). But ∠PQ G is an exterior angle for 4PQQ with remote interior angle 0 ∠PQQ . This contradicts the . Thus, we reject the RAA hypothesis and conclude that m = m0.

MTH 411/511 (Geometry) Euclid’s Elements Fall 2020 Triangle Congruence Conditions

The triangle congruence conditions allow us to check whether two triangles are congruent without going through the tedious exercise of verifying that all three pairs of angles are congruent and all three pairs of sides are congruent. We have seen one of these already, it fact is part of our . We will prove several more in the next section: • Side-Angle-Side (SAS) – VI • Angle-Side-Angle (ASA) – Theorem 4.2.1 • Angle-Angle-Side (AAS) – Theorem 4.2.3 • Hypotenuse-Leg (HL) – Theorem 4.2.5, for right triangles only • Side-Side-Side (SSS) – Theorem 4.2.7 • There is no Angle-Side-Side – Exercise 4.2.3 The ASA condition allows us to prove the following.

Converse to the Isosceles Triangle Theorem (Theorem 4.2.2) ∼ ∼ Let 4ABC is a triangle such that ∠ABC = ∠ACB, then AB = AC.

MTH 411/511 (Geometry) Euclid’s Elements Fall 2020 Next time

Before next class: Read Section 4.2.

In the next lecture we will: • Prove several triangle congruence conditions. • Prove the converse to the Isosceles Triangle Theorem.

MTH 411/511 (Geometry) Euclid’s Elements Fall 2020