Sec. 11.5 Vector methods in geometry 195
wherep+q+r=1, can be deduced from this. For
δF(Z 1 ,Z 2 ,pZ 4 +qZ 5 +rZ 6 )=δF(
Z 1 ,Z 2 ,pZ 4 +( 1 −p)(
q
1 −pZ 5 +
r
1 −pZ 6
))
=pδF(Z 1 ,Z 2 ,Z 4 )+( 1 −p)δF(
Z 1 ,Z 2 ,
q
1 −pZ 5 +
r
1 −pZ 6
)
=pδF(Z 1 ,Z 2 ,Z 4 )+( 1 −p)[
q
1 −pδF(Z 1 ,Z 2 ,Z 5 )+r
1 −pδF(Z 1 ,Z 2 ,Z 6 )]
In this we have used the fact that
q
1 −p+
r
1 −p=
q+r
1 −p=
1 −p
1 −p= 1.
11.5 Vector methods in geometry......................
There is an informative account of many of the results of this chapter contained in
Coxeter and Greitzer [5], dealt with by the methods of pure geometry.
Some results are very basic, involving just collinearities or concurrencies, or ratio
results. We start by showing how vector notation can be used to prove such results in
a very straightforward fashion.
11.5.1 Menelaus’ theorem, c. 100A.D. ....................
On using the notation of sensed
distances and sensed ratios in
7.6.1, we note that for non-
collinear points Z 1 ,Z 2 and Z 3 ,
let Z 4 ∈Z 2 Z 3 ,Z 5 ∈Z 3 Z 1 and
Z 6 ∈Z 1 Z 2 .ThenZ 4 ,Z 5 and Z 6
are collinear if and only if
Z 2 Z 4
Z 4 Z 3Z 3 Z 5
Z 5 Z 1
Z 1 Z 6
Z 6 Z 2
=− 1.
Proof.LetZ 4 =( 1 −r)Z 2 +rZ 3 ,
Z 5 =( 1 −s)Z 3 +sZ 1 ,
Z 6 =( 1 −t)Z 1 +tZ 2.
Z 1
Z 2 Z 4 Z 3
Z 5
Z 6
Figure 11.6.SinceZ 4 ,Z 5 andZ 6 are collinear, we have thatZ 6 =( 1 −u)Z 4 +uZ 5 , for some real
numberu.Then
( 1 −t)Z 1 +tZ 2 =( 1 −u)[( 1 −r)Z 2 +rZ 3 ]+u[( 1 −s)Z 3 +sZ 1 ].