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.