Geometry with Trigonometry

Sec. 11.5 Vector methods in geometry 201

and since the pointsZ 4 ,Z 5 ,Z 6 are not collinear we can equate the coefficients in the

second line, and obtain that

qt= 1 −s,( 1 −t)( 1 −m)=s,( 1 −t)m+t( 1 −q)= 0.

Now forZ 2 Z 5 andZ 3 Z 6 to meetZ 1 Z 4 in the same point, we need to have

1 −r

s−r

=−

p

q−p

,

and from this

1 −r

p

=−

s−r

q−p

But we have from above

1 −r

p

=

1 −s

q

,

as a common value oft, and so need

1 −s

q

=−

s−r

q−p

or equivalentlyq( 1 −r)=p( 1 −s), and we have already noted that this is so.

It follows thatZ 1 Z 4 ,Z 2 Z 5 andZ 3 Z 6 are concurrent.

11.5.4 Pappus’ theorem, c. 300A.D. .....................

Let the points Z 1 ,Z 2 ,Z 3 lie on
one line, and the points Z 4 ,Z 5 ,Z 6
lie on a second line, these two
lines intersecting at some point
Z 0. Suppose that Z 2 Z 6 and Z 5 Z 3
meet at W 1 ,Z 3 Z 4 and Z 6 Z 1
meet at W 2 , and Z 1 Z 5 and Z 4 Z 2
meet at W 3. Then the points
W 1 ,W 2 ,W 3 are collinear.

Z 1

Z 2

Z 3

Z 4 Z 5 Z 6

Figure 11.9.

Proof.Wehavethat

Z 2 =( 1 −p)Z 0 +pZ 1 ,Z 3 =( 1 −q)Z 0 +qZ 1 ,

Z 5 =( 1 −u)Z 0 +uZ 4 ,Z 6 =( 1 −v)Z 0 +vZ 4 ,

for somep,q,u,v∈R. On eliminatingZ 0 from the equations forZ 2 andZ 5 ,wefind

that

( 1 −u)Z 2 −( 1 −p)Z 5 =p( 1 −u)Z 1 −u( 1 −p)Z 4 ,

and so

1 −u

1 −pu

Z 2 +

u( 1 −p)

1 −pu

Z 4 =

p( 1 −u)

1 −pu

Z 1 +

1 −p

1 −pu

Z 5.