Geometry with Trigonometry

Sec. 11.5 Vector methods in geometry 203

δF(W 1 ,W 2 ,W 3 )is equal to

δF

(

q(v−u)

qv−pu

Z 2 +

u(q−p)

qv−pu

Z 6 ,

q( 1 −v)

1 −qv

Z 1 +

1 −q

1 −qv

Z 6 ,

pu( 1 −v)

v( 1 −pu)

Z 1 +

v−u

v( 1 −pu)

Z 2 +

u( 1 −p)

v( 1 −pu)

Z 6

)

=

q(v−u)

qv−pu

q( 1 −v)

1 −qv

u( 1 −p)

v( 1 −pu)

δF(Z 2 ,Z 1 ,Z 6 )

+

q(v−u)

qv−pu

1 −q

1 −qv

pu( 1 −v)

v( 1 −pu)

δF(Z 2 ,Z 6 ,Z 1 )

+

u(q−p)

qv−pu

q( 1 −v)

1 −qv

v−u

v( 1 −pu)

δF(Z 6 ,Z 1 ,Z 2 )

=

qu( 1 −v)(v−u)

v(qv−pu)( 1 −qv)(( 1 −pu)

[−q( 1 −p)+p( 1 −q)+q−p]δF(Z 1 ,Z 2 ,Z 6 )

= 0.

This shows thatW 1 ,W 2 andW 3 are collinear. This is known asPappus’ theorem.

11.5.5 Centroid of a triangle .......................

IfZ 4 ,Z 5 ,Z 6 are the mid-points of{Z 2 ,Z 3 },{Z 3 ,Z 1 },{Z 1 ,Z 2 }, respectively, then with
the notation of 11.5.2 we have thatr=s=t=^12 , and the condition (11.5.1) in the
converse of Ceva’s theorem holds. Note that[Z 3 ,Z 6 ]is a cross-bar for the interior
regionIR(|Z 3 Z 2 Z 1 )and so[Z 2 ,Z 5 meets[Z 3 ,Z 6 ]in a pointZ 7 , which is thus on
bothZ 2 Z 5 andZ 3 Z 6. It follows that it is also onZ 1 Z 4 .Thus the lines joining the
vertices of a triangle to the mid-points of the opposite sides are concurrent.The
point of concurrenceZ 7 is called thecentroidof the triangle, and for it by 11.5.2 we
have
Z 7 =^13 Z 1 +^13 Z 2 +^13 Z 3. (11.5.3)

11.5.6 Orthocentre of a triangle .......................

LetZ 8 ,Z 9 ,Z 10 be the feet of the perpendiculars fromZ 1 toZ 2 Z 3 ,Z 2 toZ 3 Z 1 ,Z 3 to
Z 1 Z 2 , respectively. Then with the notation of 11.5.2 we have that

r=

c

a

cosβ,s=

a

b

cosγ,t=

b

c

cosα.

Hence

1 −r=

a−ccosβ

a

=

ccosβ+bcosγ−ccosβ

a

=

b

a

cosγ.

Similarly

1 −s=

c

b

cosα, 1 −t=

a

c

cosβ.

It follows that the condition (11.5.1) in the converse of Ceva’s theorem is true.