Geometry with Trigonometry

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.