Geometry with Trigonometry

Sec. 3.3 A ratio result 39

(ii)For all A,B∈Π,mp(A,B)∈[A,B].

(iii)In all cases mp(A,A)=A, and mp(A,B)=A, mp(A,B)=BwhenA=B.

(iv)Given any points P and Q inΠ, there is a unique point R∈Πsuch that Q=

mp(P,R).

Proof.
(i) WhenA=Bthis follows from the definition and A 4 (ii); whenA=Bit is
immediate.
(ii) WhenA=Bthis follows from the preparatory result. WhenA=Bit amounts
toA∈{A}.
(iii) This follows from the definition and preparatory result.
(iv)Existence.IfQ=Pwe takeR=Pandthenmp(P,R)=mp(P,P)=P=Q.
Suppose then thatP=Q,letl=PQand let≤lbe the natural order onlunder which
P≤lQ.TakeRonlso thatP≤lRand|P,R|= 2 |P,Q|.ThenPprecedes bothQandR
onl, while|P,Q|=^12 |P,R|. By our initial specification ofXin the preparatory result
we see thatQ=mp(P,R).
Uniqueness. Suppose that alsoQ=mp(P,S). We again first takeQ=P.Nowin
the preparatory result we hadX=A, so that cannot be the situation here asQ=P;
thus we must haveS=Pand soS=R. Next suppose thatQ=P; then we cannot
haveS=P,aswehadX=A.ThenQ∈PS,sobyA 1 S∈PQ. In factQ∈[P,S]so
asP≤lQwe must haveP≤lS; moreover|P,R|=|P,S|as each is twice the distance
|P,Q|. By the uniqueness in A 4 (iv) we must then haveR=S.

3.3 Aratioresult..............................

3.3.1

Let A,B,C be distinct collinear points, and write

|A,C|

|A,B|

=r,

|A,C|

|C,B|

=s.

Then if C∈[A,B]we have

s=

r

1 −r

,r=

s

1 +s

.

Proof.Let|A,C|=x,|C,B|=y.AsC∈[A,B]we have|A,B|=x+y.Then

x

x+y

=r so that

x+y

x

=

1

r

.

Hence
y
x

=^1

r

−1andsox

y

=|A,C|

|C,B|

= r

1 −r

.