Geometry with Trigonometry

76 The parallel axiom; Euclidean geometry Ch. 5

We cannot haveA=Tas then we would haveA∈BD; nor canTbe any other
point of the lineAB, for then we would haveAT=ABand soC∈AB. ThusT∈AB
and similarlyTis not on any of the other side-lines.

AsB,D∈H 1 andT∈[B,D]we haveT∈H 1 .ThenA∈AB,T∈H 1 andT=A
so by 2.2.3(iv)[A,T ⊂H 1 .NowA,T,Care distinct and collinear andT∈[A,C]so
by 2.1.4(v) we cannot haveA∈[T,C]. ThusC∈[A,T and soC∈H 1. By a similar
argument the other cases follow.

We now establish a converse of this, starting withA,B,C,Das in 2.4.4, with no
three of them collinear andH 1 ,H 3 ,H 5 ,H 7 as defined there. We now suppose that
C∈H 1 ,D∈H 3 ,A∈H 5 ,B∈H 7 and wish to prove that[A,C]∩[B,D]=0./

To start with we note thatH 1 ∩H 3 =IR(|ABC)and[A,C]isacross-barof
this. Now we are given thatD∈H 1 ,D∈H 3 soDis in this interior region and by
the cross-bar theorem in 3.4.1[B,D meets[A,C]in a pointT. SimilarlyH 1 ∩H 7 =
IR(|BAD)and[B,D]is a cross-bar of this. ThusCis in this interior region and by
the cross-bar theorem[A,Cmeets[B,D]in some point. But the linesACandBDcan
meet only once as otherwise we would haveAC=BDwhich is ruled out. Thus we
haveT∈[B,D]as well.

Definition. For an integern≥3letP 1 ,P 2 ,...,Pnbenpoints such that no three of
them are collinear. For each integerjsuch that 1≤j≤n−1letH 2 j− 1 ,H 2 jbe the
closed half-planes with common edge the linePjPj+ 1 ,andH 2 n− 1 ,H 2 nthe closed
half-planes with edgePnP 1. Suppose that all the pointsPklie inH 2 j− 1 in each case.
Then the intersection

⋂n
j= 1 H^2 j−^1 is called aconvex polygon. The intersection of
the corresponding open half-planes is called theinteriorof the convex polygon. The
notation for convex quadrangles is extended to convex polygons in a straightforward
way.

Consider a convex polygonal region with sides[P 1 ,P 2 ],[P 2 ,P 3 ],...,[Pn,P 1 ].Leta
point U interior to the polygon be joined by segments to the vertices. Then

n− 1

∑

j= 1

Δ[U,Pj,Pj+ 1 ]+Δ[U,Pn,P 1 ]=

n− 1

∑

j= 2

Δ[P 1 ,Pj,Pj+ 1 ]. (5.6.1)

Proof. CASE 1. We first take the case of a triangle so thatn=3. Now[P 1 ,U will
meet[P 2 ,P 3 ]in a pointV. Then by 5.6.1(i)

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

=Δ[U,P 1 ,P 2 ]+{Δ[U,P 2 ,V]+Δ[U,V,P 3 ]}+Δ[U,P 3 ,P 1 ]

={Δ[U,P 1 ,P 2 ]+Δ[U,P 2 ,V]}+{Δ[U,V,P 3 ]+Δ[U,P 3 ,P 1 ]}

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