Geometry with Trigonometry

78 The parallel axiom; Euclidean geometry Ch. 5

[P 1 ,U meets[P 2 ,P 3 ]at a pointV=P 2 so thatU∈[P 1 ,V].Also[P 1 ,U meets[Pn− 1 ,Pn]
at a pointW=Pnso thatU∈[P 1 ,W]. If we writel=P 1 Uand haveP 1 ≤lU,then
U≤lV,U≤lWand asV≤lWorW≤lVwe haveV∈[U,W]orW∈[U,V].
One possibility is thatW=V. We cannot have eitherPn− 1 orPnonP 2 P 3 as that
would make three pointsPkcollinear. Then we have thatV=W∈[Pn− 1 ,Pn]and
V∈P 2 P 3 so thatPn− 1 andPnare on different sides ofP 2 P 3 , that is one is inH 3 and
the other is not, whereas all the verticesPkare inH 3. This contradiction implies that
V=W.
A second possibility is thatV∈[U,W],V=W.ThenW∈H 3 whereasPn− 1 ,Pn∈
H 3 and asW∈[Pn− 1 ,Pn]⊂H 3 we have a contradiction.
A third and final possibility is thatW∈[U,V],W=V.ThenV∈H 2 n− 3 whereas
P 2 ,P 3 ∈H 2 n− 3 and asV∈[P 2 ,P 3 ]⊂H 2 n− 3 we have a contradiction again.
With this preparation suppose now thatU∈[P 1 ,P 2 ,P 3 ]. By 5.6.1(ii)

Δ[U,P 1 ,P 2 ]+Δ[U,P 2 ,P 3 ]=Δ[P 1 ,P 2 ,P 3 ]+Δ[U,P 1 ,P 3 ].

Hence asUis interior to the polygon withnsides[P 1 ,P 3 ],[P 3 ,P 4 ], ...,

[Pn,Pn+ 1 ],[Pn+ 1 ,P 1 ],

n

∑

j= 1

Δ[U,Pj,Pj+ 1 ]+Δ[U,Pn+ 1 ,P 1 ]

=Δ[P 1 ,P 2 ,P 3 ]+Δ[U,P 1 ,P 3 ]+

n

∑

j= 3

Δ[U,Pj,Pj+ 1 ]+Δ[U,Pn+ 1 ,P 1 ]

=Δ[P 1 ,P 2 ,P 3 ]+

n

∑

j= 3

Δ[P 1 ,Pj,Pj+ 1 ]=

n

∑

j= 2

Δ[P 1 ,Pj,Pj+ 1 ].

If insteadU∈[P 1 ,Pn,Pn+ 1 ]we get the same conclusion by similar reasoning. The

result now follows by induction onn.

Definition.Theareaof the polygonal region in the present section is defined to

be the sum of the areas of the triangles involved, as in (5.6.1).

Exercises

5.1 Opposite wedge-angles in a parallelogram have equal degree-measures.

5.2 If two adjacent sides of a rectangle have equal lengths, then all the sides have

equal lengths. Such a rectangle is called asquare.

5.3 If the diagonals of a parallelogram have equal lengths, it must be a rectangle.

5.4 If the diagonal lines of a rectangle are perpendicular, it must be a square.

5.5 LetA,B,Cbe non-collinear points and letP∈[A,BandQ∈[A,Cbe such that

|A,P|

|A,B|

=

|A,Q|

|A,C|

.