Sec. 11.5 Vector methods in geometry 199
11.5.3 Desargues’ perspective theorem, 1648 A.D. ...........
Let(Z 1 ,Z 2 ,Z 3 )and(Z 4 ,Z 5 ,Z 6 )
be two pairs of non-collinear
points. Let Z 2 Z 3 and Z 5 Z 6 meet
at W 1 ,Z 3 Z 1 and Z 6 Z 4 meet at W 2 ,
and Z 1 Z 2 and Z 4 Z 5 meet at W 3.
Then W 1 ,W 2 ,W 3 are collinear if
and only if Z 1 Z 4 ,Z 2 Z 5 ,Z 3 Z 6 are
concurrent.
Proof. Suppose thatZ 1 Z 4 ,Z 2 Z 5 ,
Z 3 Z 6 meet at a pointZ 0 .Then
Z 4 =( 1 −u)Z 0 +uZ 1 ,
Z 5 =( 1 −v)Z 0 +vZ 2 ,
Z 6 =( 1 −w)Z 0 +wZ 3 ,for someu,v,w,∈R.
Z 1
Z 2
Z 3
Z 4
Z 5
Z 6
W 1
W 2
W 3
Figure 11.8.On eliminatingZ 0 between the second and third of these, we obtain that( 1 −w)Z 5 −( 1 −v)Z 6 =v( 1 −w)Z 2 −w( 1 −v)Z 3 ,from which we obtain that
1 −w
v−wZ 5 −
1 −v
v−wZ 6 =
v( 1 −w)
v−wZ 2 −
w( 1 −v)
v−wZ 3.
Now the sum of the coefficients ofZ 5 andZ 6 is equal to 1, so the left-hand side
represents a point on the lineZ 5 Z 6. Similarly, the sum of the coefficients ofZ 2 and
Z 3 is equal to 1, so the right-hand side represents a point on the lineZ 2 Z 3. Thus this
must be the pointW 1.
By a similar argument based on the third and first lines, we find that1 −u
w−uZ 6 −
1 −w
w−uZ 4 =
w( 1 −u)
w−uZ 3 −
u( 1 −w)
w−uZ 1
must be the pointW 2 , and by a similar argument based on the first and second lines,
we find that
1 −v
u−vZ 4 −
1 −u
u−vZ 5 =
u( 1 −v)
u−vZ 1 −
v( 1 −u)
u−vZ 2
must be the pointW 3.