Geometry with Trigonometry

66 The parallel axiom; Euclidean geometry Ch. 5

Let A,B,C be non-collinear points and let P∈[A,B and Q∈[A,C be such that
PQ‖BC. Then
|A,P|
|A,B|

=

|A,Q|

|A,C|

.

A

B C

P Q

B 1

B 2

B 4

B 5

B 6

B 7

B 8

B 9

C 1

C 2

C 4

C 5

C 6

C 7

C 8

C 9

Figure 5.6.

A

B C

P Q

Pu

Pv

Qu

Qv

Proof. We assume first thatP∈[A,B]. Within this first case, we suppose initially
that
|A,P|
|A,B|

=

s

t

,

wheresandtare positive whole numbers withs<t,sothats/tis an arbitrary rational
number between 0 and 1. For 0≤j≤tletBjbe the point on[A,B such that

|A,Bj|

|A,B|

=

j

t

,

so thatB 0 =A,Bt=BandBs=P.IfAB=land≤lis the natural order for which
A≤lB,thenA≤lBj− 1 ≤lBj≤lBj+ 1 ≤lBand|Bj− 1 ,Bj|=|Bj,Bj+ 1 |.IfAC=m
and≤mis the natural order for whichA≤mC, then by the last result applied with
(D 1 ,D 2 ,D 3 )=(Bj− 1 ,Bj,Bj+ 1 ),for1≤j≤t−1 the line throughBjwhich is parallel
toBCwill meetACin a pointCjsuch thatA≤mCj− 1 ≤mCj≤mCj+ 1 ≤mCand
|Cj− 1 ,Cj|=|Cj,Cj+ 1 |.
It follows that, for 0≤j≤t,|A,Cj|=j|A,C 1 |andsoasCt=C,

|A,Cj|

|A,C|

=

j|A,C 1 |

t|A,C 1 |

=

j

t

.

In particular, asCs=Q, it follows that

|A,Q|

|A,C|

=

s

t

=

|A,P|

|A,B|

.