200 Vector and complex-number methods Ch. 11
Then by repeated use of 10.5.3 and 11.4.5
δF(W 1 ,W 2 ,W 3 )
=δF
(
v( 1 −w)
v−w
Z 2 −
w( 1 −v)
v−w
Z 3 ,−
u( 1 −w)
w−u
Z 1 +
w( 1 −u)
w−u
Z 3 ,
u( 1 −v)
u−v
Z 1 −
v( 1 −u)
u−v
Z 2
)
=
[
v( 1 −w)
v−w
w( 1 −u)
w−u
u( 1 −v)
u−v
−
w( 1 −v)
v−w
−u( 1 −w)
w−u
−v( 1 −u)
u−v
]
δF(Z 1 ,Z 2 ,Z 3 )
= 0.
This shows thatW 1 ,W 2 andW 3 are collinear.
This is known asDesargues’perspective or two-triangle theorem.
Conversely, let
W 1 =( 1 −l)Z 2 +lZ 3 =( 1 −m)Z 5 +mZ 6 ,
W 2 =( 1 −p)Z 3 +pZ 1 =( 1 −q)Z 6 +qZ 4 ,
W 3 =( 1 −r)Z 1 +rZ 2 =( 1 −s)Z 4 +sZ 5.
From the third of these we deduce that( 1 −r)Z 1 −( 1 −s)Z 4 =sZ 5 −rZ 2 , and from
this
1 −r
s−r
Z 1 −
1 −s
s−r
Z 4 =
s
s−r
Z 5 −
r
s−r
Z 2 ,
so that this must be the point of intersection ofZ 1 Z 4 andZ 2 Z 5.
By a similar argument, we deduce from the second equation that
1 −l
m−l
Z 2 −
1 −m
m−l
Z 5 =
m
m−l
Z 6 −
l
m−l
Z 3 ,
and so this must be the point of intersection ofZ 2 Z 5 andZ 3 Z 6. By a similar argument,
we deduce from the first equation that
1 −p
q−p
Z 3 −
1 −p
q−p
Z 6 =
q
q−p
Z 4 −
p
q−p
Z 1 ,
and so this must be the point of intersection ofZ 3 Z 6 andZ 1 Z 4.
We are given now thatW 1 ,W 2 andW 3 are collinear, so thatW 3 =( 1 −t)W 1 +tW 2 ,
for somet∈R.Then
( 1 −t)[( 1 −l)Z 2 +lZ 3 ]+t[( 1 −p)Z 3 +pZ 1 ]=( 1 −r)Z 1 +rZ 2 ,
( 1 −t)[( 1 −m)Z 5 +mZ 6 ]+t[( 1 −q)Z 6 +qZ 4 ]=( 1 −s)Z 4 +sZ 5.
Since the pointsZ 1 ,Z 2 ,Z 3 are not collinear we can equate the coefficients in the first
line here, and obtain that
pt= 1 −r,( 1 −t)( 1 −l)=r,( 1 −t)l+t( 1 −p)= 0 ,