Geometry with Trigonometry

(Marvins-Underground-K-12) #1

36 Distance; degree-measure of an angle Ch. 3


contains infinitely many points and this supersedes the specification in A 1 thatl=0;/
nevertheless it was convenient to stipulate the latter to avoid a trivial situation. In
A 4 (iii) addition + of real numbers is involved.





P


Q


R


x

y

x+

y

Figure 3.1. Addition of distances.



A


k B

Laying off a distancek.

3.1.2 Derived properties of distance.....................


Distance has the following properties:-


(i)For all A∈Π|A,A|= 0 , and we have|A,B|> 0 if A=B.

(ii) If P∈[A,B],then|A,P|≤|A,B|. If additionally P=B, then|A,P|<|A,B|.

(iii) If A=C and B lies on the line AC but outside the segment[A,C],then

|A,B|+|B,C|>|A,C|.

(iv) If C∈[A,B is such that|A,B|<|A,C|,thenB∈[A,C].

Proof.
(i) By A 4 (iii) withP=Q=Aand anyR∈Π,wehave|A,A|+|A,R|=|A,R|,i.e.
x+y=ywherex=|A,A|andy=|A,R|. It follows thatx=0.
Next withA=Bletl=ABand≤lbe the natural order onlfor whichA≤lB.
Then we have
A≤lB,A≤lA,|A,A|= 0 ,


so that if we also had|A,B|=0, then by the uniqueness part of A 4 (iv) withk=0, we
would haveA=Band so have a contradiction. To avoid this we must have|A,B|>0.
(ii) AsP∈[A,B],byA 4 (iii) we have|A,P|+|P,B|=|A,B|. But by A 4 (i)|P,B|≥ 0
and so|A,P|≤|A,B|.IfP=B, then by (i) of the present theorem|P,B|>0andso
|A,P|<|A,B|.
(iii) AsB∈[A,C]we haveB=A,B=Candsoby2.1.4wehaveeitherA∈[B,C]
orC∈[A,B]. In the first of these|B,A|+|A,C|=|B,C|by A 4 (iii) and as|A,B|=
|B,A|>0thisgives|A,C|<|B,C|<|A,B|+|B,C|. In the second case we have
|A,C|+|C,B|=|A,B|by A 4 (iii) and as|C,B|=|B,C|>0, then|A,C|<|A,B|<
|A,B|+|B,C|.

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