Geometry with Trigonometry

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 ,