Sec. 5.3 Ratio results for triangles 65
Proof. By Pasch’s property in
2.4.3 applied to the triangle
[A,B,C], the lines through
[D 1 ,D 2 ,D 3 ]which are parallel
toBCwill meet[A,C]in points
E 1 ,E 2 ,E 3 , respectively. By
Pasch’s property applied to
[A,D 3 ,E 3 ],sinceD 2 ∈[A,D 3 ]
and the lines through D 2 and
D 3 parallel toBCare parallel to
each other,E 2 ∈[A,E 3 ].
By Pasch’s property applied to
[A,D 2 ,E 2 ],sinceD 1 ∈[A,D 2 ]
and the lines through D 1 and
D 2 parallel toBCare parallel to
each other,E 1 ∈[A,E 2 ].Itre-
mains to show thatE 2 is equidis-
tant fromE 1 andE 3.
|||A
B C
D 1
D 2
D 3
E 1
E 2
E 3
F
G
||xxyyFigure 5.5. Transversals to parallel lines.By Pasch’s property applied to[A,D 2 ,E 2 ],sinceE 1 ∈[A,E 2 ]the line throughE 1
which is parallel toAB=AD 2 will meet[D 2 ,E 2 ]in a pointF. By Pasch’s property
applied to[A,D 3 ,E 3 ],sinceE 2 ∈[A,E 3 ]the line throughE 2 which is parallel toAB=
AD 3 will meet[D 3 ,E 3 ]in a pointG.
LetT 1 ,T 2 be the triangles with vertices{E 1 ,F,E 2 },{E 2 ,G,E 3 }, respectively. Our
objective is to show that
T (^1) (E 1 ,F,E 2 )→≡(E 2 ,G,E 3 )T 2.
NowD 1 E 1 ‖D 2 F,D 1 D 2 ‖E 1 F,so[D 1 ,D 2 ,F,E 1 ]is a parallelogram, and so by 5.2.1
|D 1 ,D 2 |=|E 1 ,F|. SimilarlyD 2 E 2 ‖D 3 G,D 2 D 3 ‖E 2 Gso[D 2 ,D 3 ,G,E 2 ]is a parallel-
ogram, and so|D 2 ,D 3 |=|E 2 ,G|.But|D 1 ,D 2 |=|D 2 ,D 3 |and hence|E 1 ,F|=|E 2 ,G|.
LetH 1 be the closed half-plane with edgeACin whichBlies. Then[A,B]⊂
H 1 ,soD 2 ,D 3 ∈H 1 .Then[D 2 ,E 2 ],[D 3 ,E 3 ]⊂H 1 ,soF,G∈H 1 .ThenFandG
are on the one side of the lineAC,andasD 2 E 2 ‖D 3 E 3 andE 2 ∈[E 3 ,E 1 ],thean-
gles∠FE 2 E 1 ,∠GE 3 E 2 are corresponding angles for parallel lines and so have equal
degree-measures. Thus|∠FE 2 E 1 |◦=|∠GE 3 E 2 |◦.
By transitivityE 1 F‖E 2 Gas both are parallel toAB,FandGare on the one side of
AC,andE 2 ∈[E 1 ,E 3 ], so the angles∠FE 1 E 2 and∠GE 2 E 3 are corresponding angles
for parallel lines and so have equal degree-measures. Thus|∠FE 1 E 2 |◦=|∠GE 2 E 3 |◦.
As
|∠FE 2 E 1 |◦=|∠GE 3 E 2 |◦,|∠FE 1 E 2 |◦=|∠GE 2 E 3 |◦,
by 5.2.2|∠E 1 FE 2 |◦=|∠E 2 GE 3 |◦. Thus
|E 1 ,F|=|E 2 ,G|,|∠FE 1 E 2 |◦=|∠GE 2 E 3 |◦,|∠E 1 FE 2 |◦=|∠E 2 GE 3 |◦,
so by the ASA principle, the trianglesT 1 ,T 2 are congruent in the correspondence
(E 1 ,F,E 2 )→(E 2 ,G,E 3 ). It follows that|E 1 ,E 2 |=|E 2 ,E 3 |.