Geometry with Trigonometry

Sec. 11.5 Vector methods in geometry 197

From this

s

1 −t

=

1 −v

1 −w

,

t

1 −r

=

1 −w

1 −u

,

r

1 −s

=

1 −u

1 −v

,

and so by multiplication
s
1 −t

t

1 −r

r

1 −s

= 1.

Thus we obtain our conclusion. This is known asC eva’s th eo rem.
In fact we also have that
u
s

=−

1 −u

1 −v

,

v

t

=−

1 −v

1 −w

,

w

r

=−

1 −w

1 −u

,

which givesuvw=−rst.Thisis

Z 0 Z 4

Z 0 Z 1

Z 0 Z 5

Z 0 Z 2

Z 0 Z 6

Z 0 Z 3

=−

Z 2 Z 4

Z 2 Z 3

Z 3 Z 5

Z 3 Z 1

Z 1 Z 6

Z 1 Z 2

.

CONVERSEofC eva’s th eo rem.Conversely, for non-collinear points Z 1 ,Z 2
and Z 3 ,letZ 4 ∈Z 2 Z 3 ,Z 5 ∈Z 3 Z 1 and Z 6 ∈Z 1 Z 2. If (11.5.1) holds and Z 2 Z 5 and Z 3 Z 6
meet at a point Z 0 ,thenZ 1 Z 4 also passes through Z 0.
To start our proof we note that we have

Z 5 =( 1 −v)Z 0 +vZ 2 =( 1 −s)Z 3 +sZ 1 ,

Z 6 =( 1 −w)Z 0 +wZ 3 =( 1 −t)Z 1 +tZ 2.

Hence

Z 0 =

s

1 −v

Z 1 −

v

1 −v

Z 2 +

1 −s

1 −v

Z 3 ,Z 0 =

1 −t

1 −w

Z 1 +

t

1 −w

Z 2 −

w

1 −w

Z 3.

It follows that

s

1 −v

=

1 −t

1 −w

,−

v

1 −v

=

t

1 −w

,

1 −s

1 −v

=−

w

1 −w

,

from which
s
1 −t

=

1 −v

1 −w

,( 1 −s)t=vw.

On eliminatingsbetween these, we obtain( 1 −v)t^2 +(v−w)t−vw( 1 −w)=0. We
then obtain two pairs of solutions,t=w,s= 1 −v,and

t=−v

1 −w

1 −v

,s=

1 −vw

1 −w

.

The first pair of solutions leads tov=w=0andsoZ 5 =Z 6 =Z 0 =Z 1 ,whichwe
regard as a degenerate case.
WithZ 4 =( 1 −r)Z 2 +rZ 3 , we are given that
1 −r
r

=

st

( 1 −s)( 1 −t)

,