Geometry with Trigonometry

28 Basic shapes of geometry Ch. 2

CASE (b). LetA,B∈l.Then[A,B]⊂l⊂H 1 , so the result follows.
CASE (c). Let one ofA,Bbe onland the other inG 1 , sayA∈l,B∈G 1.
Suppose that[A,B]is not a subset ofH 1. Then there is some pointC∈[A,B]such
thatC∈G 2. Note thatC=A,C=BasA,B∈G 2 ,C∈G 2.
NowB∈G 1 ,C∈G 2 so byA 3 (iii) there is some pointDof[B,C]onl,sothat
D∈[B,C],D∈l.ThenA,B,Care collinear and distinct, andC∈[A,B]so by 2.1.4
we cannot haveA∈[B,C]. HenceA=D.
ButA∈l,D∈lso byA 1 ,AD=l.HoweverAB=BCandD∈BC,soD∈AB.
ThenAB=AD=l,soB∈l. This gives a contradiction. Thus the original supposition
is untenable so[A,B]⊂H 1 , and this proves (iii).
(iv)
CASE (a). LetB∈l.Then[A,B⊂l⊂H 1 , which gives the desired conclusion.
CASE (b). LetB∈G 1. Suppose that[A,Bis not a subset ofH 1. Then there is
some pointC∈[A,Bsuch thatC∈G 2. ClearlyC=A,C=B.NowA,B,Care distinct
collinear points, so by 2.1.4 precisely one of

A∈[B,C],B∈[C,A],C∈[A,B],

holds. We cannot haveA∈[B,C]as that would putB,Cin different half-lines with
initial-pointA, whereas they are both in[A,B. This leaves us with two subcases.
Subcase1. LetC∈[A,B]. We recall thatA,B∈H 1 so by part (iii) of the present
result[A,B]⊂H 1 .AsC∈[A,B],C∈G 2 , we have a contradiction.
Subcase2. LetB∈[A,C]. We recall thatA∈H 2 ,C∈H 2 so by part (iii) of the
present result,[A,C]⊂H 2 .ThenB∈H 2 ,B∈G 1 which gives a contradiction. Thus
the original supposition is untenable, and this proves (iv).

NOTE. The terms ‘open’ and ‘closed’ are standard in analysis and point-set topol-
ogy. What is significant is that an open half-plane contains none of the points of the
edge, while a closed half-plane contains all of the points of the edge.

2.3 Angle-supports, interior and exterior regions, angles

2.3.1 Angle-supports, interior regions

A

B

C

Figure 2.6. An angle-support.

A

B

C

A straight angle-support.