Geometry with Trigonometry

Sec. 5.6 Area of triangles, and convex quadrilaterals and polygons 77

P 1

P 2 V P 3

U

P 1

P 2

P 3

U

P 4

P 1

P 2 P 3

U

P 4

P 1

P 2

P 3

U

P 4

P 5

P 6

Figure 5.16.

CASE 2. Secondly we take the case of a convex quadrilateral so thatn=4. Sup-

pose first thatU∈[P 1 ,P 3 ]. Then by 5.6.1(i) used twice,

{Δ[U,P 1 ,P 2 ]+Δ[U,P 2 ,P 3 ]}+{Δ[U,P 3 ,P 4 ]+Δ[U,P 4 ,P 1 ]}

=Δ[P 1 ,P 2 ,P 3 ]+Δ[P 1 ,P 3 ,P 4 ].

Suppose next thatU∈[P 1 ,P 3 ].ThenU is interior to[P 1 ,P 2 ,P 3 ]or[P 1 ,P 3 ,P 4 ], say
U∈[P 1 ,P 3 ,P 4 ].
We need to show that we have a convex quadrilateral[P 1 ,P 2 ,P 3 ,U].NowU∈
IR(|P 1 P 2 P 3 )for which[P 1 ,P 3 ]is a cross-bar. Thus[P 2 ,U meets[P 1 ,P 3 ]in a point
T. Now we cannot haveU∈[P 2 ,T]as that would makeU∈[P 1 ,P 2 ,P 3 ], nor can we
haveP 2 ∈[T,U]as that would putUoutsideH 1 andH 3 ,sowemusthaveT∈[P 2 ,U].
Then 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 ],

so

Δ[U,P 1 ,P 2 ]+Δ[U,P 2 ,P 3 ]+Δ[U,P 3 ,P 4 ]+Δ[U,P 4 ,P 1 ]

=Δ[P 1 ,P 2 ,P 3 ]+{Δ[U,P 1 ,P 3 ]+Δ[U,P 3 ,P 4 ]+Δ[U,P 4 ,P 1 ]}

=Δ[P 1 ,P 2 ,P 3 ]+Δ[P 1 ,P 3 ,P 4 ]

by CASE 1.

CASE 3. We now suppose that the result holds, for somen≥4, for any convex

polygonal region withnsides. Then for thatnconsider any convex polygonal region

withn+1 sides,[P 1 ,P 2 ],[P 2 ,P 3 ], ...,[Pn,Pn+ 1 ],[Pn+ 1 ,P 1 ].

Asn+ 1 ≥ 5 ,[P 1 ,P 2 ,P 3 ]and[P 1 ,Pn,Pn+ 1 ]have onlyP 1 in common, soUcannot

be in both. For suppose thatUis an interior point which is common to both. Then