196 Vector and complex-number methods Ch. 11
As the coefficients on each side add to 1, by the uniqueness in 11.4.2 we can equate
coefficients and thus obtain
1 −t=su,t=( 1 −u)( 1 −r),r( 1 −u)=−u( 1 −s).
On eliminatinguwe obtain
r
1 −s
s
1 −t
t
1 −r
=−
u
1 −u
1
u
( 1 −u)=− 1 ,
and so r
1 −r
s
1 −s
t
1 −t
=− 1.
This yields the stated result. This is known asMenelaus’ theorem.
11.5.2 Ceva’s theorem and converse, 1678 A.D. ...............
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
Z 1 Z 4 ,Z 2 Z 5 and Z 3 Z 6 are concurrent, then
Z 2 Z 4
Z 4 Z 3
Z 3 Z 5
Z 5 Z 1
Z 1 Z 6
Z 6 Z 2
= 1. (11.5.1)
Proof. Denoting the point of concurrency byZ 0 ,wehave
Z 4 =( 1 −u)Z 0 +uZ 1 =( 1 −r)Z 2 +rZ 3 ,
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 ,
for someu,v,w,r,s,t∈R.Then
Z 0 =−
u
1 −u
Z 1 +
1 −r
1 −u
Z 2 +
r
1 −u
Z 3 ,
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.
On equating the coefficients of
Z 1 ,Z 2 andZ 3 , in turn, we obtain
−
u
1 −u
=
s
1 −v
=
1 −t
1 −w
,
1 −r
1 −u
=− v
1 −v
= t
1 −w
,
r
1 −u
=
1 −s
1 −v
=−
w
1 −w
.
Z 1
Z 2 Z 4 Z 3
Z (^6) Z 5
Figure 11.7.