Geometry with Trigonometry

Sec. 3.6 Mid-line of an angle-support 45

3.6.3 Mid-lines................................

Given any angle-support|BACsuch that C∈[A,B , there is a unique line l such that
A∈l and for all P=AbutP∈l,|∠BAP|◦=|∠PAC|◦.
Proof.Existence.
This was already shown in 3.6.1 in the case when|BACis straight, so we may
assume thatA,B,Care non-collinear.
By A 5 (v) and 3.5.2, as 0<|∠BAC|◦<180andso0<^12 |∠BAC|◦<90, there is a
half-line[A,PwithPon the same side ofABasCis, such that|∠BAP|◦=^12 |∠BAC|◦.
Then[A,P⊂IR(|BAC)by 3.5.2, so by A 5 (iii)

|∠BAP|◦+|∠PAC|◦=|∠BAC|◦.

A

B

C

P

B 1

H 1

H 4 H 3

A B

C P

B 1

H 1

H 4 H 3

Figure 3.11.
It follows that

|∠PAC|◦=|∠BAC|◦−

1

2

|∠BAC|◦=

1

2

|∠BAC|◦

and so|∠BAP|◦=|∠PAC|◦.
IfP′=Ais such thatA∈[P,P′],thenbyA 5 (iii)

|∠BAP′|◦= 180 −|∠BAP|◦= 180 −|∠PAC|◦=|∠P′AC|◦.

Uniqueness.
When|BACis straight, by A 5 (iii) 2|∠BAP|◦=180 so|∠BAP|◦=90. By A 5 (iv)
this determinesluniquely. For the remainder we suppose then that we have a wedge-
angle∠BAC.
LetH 1 ,H 2 be the closed half-planes with common edgeAB, withC∈H 1 ,and
H 3 ,H 4 be the closed half-planes with common edgeAC, withB∈H 3 .LetB 1 =A
be such thatA∈[B,B 1 ].
Now iflcontains a pointQ=AinH 4 it will also contain a pointR=AofH 3 ,
so we may assume thatlcontains a pointP=AofH 3 .As

H 1 =H 1 ∩Π=H 1 ∩(H 3 ∪H 4 )=(H 1 ∩H 3 )∪(H 1 ∩H 4 )

=IR(|BAC)∪IR(|B 1 AC),

we then haveP∈IR(|BAC)orP∈IR(|B 1 AC).
We get a contradiction iflis eitherABorAC.Forifl=AB,thenwehave
|∠BAP|◦= 0 ,|∠PAC|◦>0. Similarly ifl=AC.