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Lecture 20: Exterior Angle Theorem
20.1 Comparing segments and angles
Definition In a metric geometry, we say AB is less than, or smaller than, CD, denoted AB < CD, if AB < CD, AB is greater than, or larger than CD, denoted AB > CD, if AB > CD, and we write AB ≤ CD if either AB < CD or AB ' CD.
Definition In a protractor geometry, we say ∠ABC is less than, or smaller than, ∠DEF , denoted ∠ABC < ∠DEF , if m(∠ABC) < m(∠DEF ), ∠ABC is greater than, or larger than ∠DEF , denoted ∠ABC > ∠DEF , if m(∠ABC) > m(∠DEF ), and we write ∠ABC ≤ ∠DEF if either ∠ABC < ∠DEF or ∠ABC ' ∠DEF.
Theorem In a metric geometry, AB < CD if and only if there exists a point G ∈ int(CD) such that AB ' CG.
Theorem In a protractor geometry, ∠ABC < ∠DEF if and only if there exists a point G ∈ int(∠DEF ) such that ∠ABC ' ∠DEG.
20.2 Exterior angles
Definition Given 4 ABC in a protractor geometry and a point D with A − C − D, we call ∠BCD an exterior angle of 4 ABC with remote interior angles ∠BAC and ∠ABC.
Exterior Angle Theorem In a neutral geometry, an exterior angle of 4 ABC is greater than either of its remote interior angles.
Proof Given 4 ABC, let D be a point with A−C −D. We will first show that ∠BCD > ∠ABC.
Let M be the midpoint of AB and let E be a point on
−→ AM with A − M − E and AM ' M E. Then AM ' EM , ∠AM C ' ∠EM C (they are vertical angles), and M B ' M C, and so 4 AM C ' 4EM C by Side-Angle-Side. In particular, ∠ABC ' ∠BCE. Since E ∈ int(∠BCD) (this was a homework problem), it follows that
∠BCD > ∠BCE ' ∠ABC,
and so ∠BCD > ∠ABC.
It remains to show that ∠BCD > ∠BAC. If we let D′^ be a point on
−→ BC with B − C − D′, then, by what we have already shown, ∠ACD′^ > ∠BAC. Since ∠ACD′^ ' ∠BCD (see homework), it follows that ∠BCD > ∠BAC.
Lecture 20: Exterior Angle Theorem 20-
Theorem Given a point P and a line in a neutral geometry, there exists a unique line through P perpendicular to.
Proof We have already proven the result for P ∈ . If P ∈/, we have already shown there exists at least one line through P perpendicular to `.
Now suppose lines m and n are both perpendicular to . Let {A} = ∩ m, {B} = ∩ n, and let C be a point on with A − B − C. Then ∠P BC is an exterior angle of 4 AP B, and so ∠P BC > ∠P AB, contradicting the assumption that both ∠P AB and ∠P BC are right angles (and hence congruent).
Side-Angle-Angle (SAA) If, in a neutral geometry, triangles 4 ABC and 4 DEF are such that AB ' DE, ∠A ' ∠D, and ∠C ' ∠F , then 4 ABC ' 4DEF.
Proof If AC ' DF , then 4 BAC ' 4EDF by Side-Angle-Side. Thus 4 ABC ' 4 DEF.
So suppose AC < DF. Let G ∈ DF so that AC ' DG. Then 4 BAC ' 4EDG by Side-Angle-Side. In particular, ∠ACB ' ∠DGE. However, ∠DGE is an exterior angle of 4 EGF ; in particular, ∠DGE > ∠GF E = ∠DF E. Thus we have ∠ACB > ∠DF E, contradicting our assumptions. Thus we cannot have AC < DF. Similarly, we cannot have DF < AC, and so AC ' DF , and 4 ABC ' 4DEF.
Note: Another approach to Side-Angle-Angle would be to argue that ∠B ' ∠E and then conclude 4 ABC ' 4DEF by Side-Angle-Side. Why can’t we use this approach?
Theorem Given 4 ABC in a neutral geometry, if AB > AC, then ∠C > ∠B.
Proof Let D ∈
−→ AC be such that A − C − D and AD ' AB. Then 4 ABC is isosceles, and so ∠ADB ' ∠ABD. Moreover, C ∈ int(∠ABD), so ∠ABD > ∠ABC. Since ∠ACB is an exterior angle of 4 BCD, we now have
∠ABC < ∠ABD ' ∠ADB < ∠ACB.
Theorem Given 4 ABC in a neutral geometry, if ∠C > ∠B, then AB > AC.
Proof See homework.
Triangle Inequality Given 4 ABC in a neutral geometry, AC < AB + BC.
Proof Let D ∈
−→ BC such that C − B − D and BD ' AB. Then 4 ABD is isosceles, so ∠BAD ' ∠ADB. Now B ∈ int(∠DAC), so ∠BAD < ∠CAD. Hence, in 4 ACD, ∠ADC < ∠CAD, and so
AC < DC = CB + BD = BC + AB.
