Sec. 5.5 Mid-lines and triangles 71
It follows that|∠EAC|◦=|∠ACH|◦,andasE,Hare on opposite sides ofAC this
implies thatAE‖HC. It then follows that
|B,E|
|E,C|=
|B,A|
|A,H|
=
|B,A|
|A,C|
.
Next chooseK∈[A,B so that|A,K|=|A,C|.Then|A,K|<|A,B|soK∈[A,B].
Now
|∠HAC|◦=|∠AKC|◦+|∠ACK|◦, |∠AKC|◦=|∠ACK|◦,
|∠HAC|◦=|∠HAG|◦+|∠GAC|◦, |∠HAG|◦=|∠GAC|◦.It follows that|∠GAC|◦=|∠ACK|◦.ButH,Kare on opposite sides ofAC,H,Gare
on the same side, and soG,Kare on opposite sides. This implies thatAG‖KC.Now
AGmeetsBCatF,andK∈[A,B]soC∈[B,F]. It follows that
|B,F|
|F,C|=
|B,A|
|A,K|
=
|B,A|
|A,C|
.
On combining the two results, we then have
|B,E|
|E,C|=|B,F|
|F,C|
.
NOTE. We also refer to the mid-line of|CADabove as theexternal bisectorof
|BAC.When{E,F}divide[B,C]internally and externally in the same ratio, we say
that(B,C,E,F)form aharmonic range.
Let(A,B,C,D)be a harmonic range and S∈AB. Let the line through C, parallel
to SD, meet SA at G and SB at H. Then C is the mid-point of G and H.
A
B
C
D
S
G
H Figure 5.11.
Proof.Wearegiven
|A,C|
|C,B|=
|A,D|
|D,B|
,
so
|A,C|
|A,D|