204 Vector and complex-number methods Ch. 11
Repeating an argument that we
used in 7.2.3, suppose now that
m and n are any lines which
are perpendicular to Z 3 Z 1 and
Z 1 Z 2 , respectively. If we hadm‖
n, then we would have m⊥
Z 3 Z 1 ,n‖mso thatn⊥Z 3 Z 1 ;but
alreadyn⊥Z 1 Z 2 soZ 3 Z 1 ‖Z 1 Z 2 ;
as Z 1 ,Z 2 ,Z 3 are not collinear,
this gives a contradiction.
Z 1
Z 2
Z 3
Z 11
Z 8
Z 9
Z 10
Figure 11.10. Orthocentre of triangle.
Thusmis not parallel tonand so these lines meet in a unique pointZ 11. In partic-
ular the linesZ 2 Z 9 ,Z 3 Z 10 must meet in a unique pointZ 11 and then by the converse of
Ceva’s theorem,Z 1 Z 4 will pass throughZ 11 .Thus the lines through the vertices of a
triangle which are perpendicular to the opposite side-lines are concurrent.The point
of concurrenceZ 11 is called theorthocentreof the triangle.
By (11.5.2) we thus have
Z 11 =
a
bcosγ
a
ccosβ
a
ccosβ+
a
ccosγcosα
Z 1 +
a
ccosγcosα
a
ccosβ+
a
ccosγcosα
Z 2 +
a
bcosαcosβ
a
ccosβ+
a
ccosγcosα
Z 3
=
a
b
cosβcosγ
cosβ+cosγcosα
Z 1 +
cosγcosα
cosβ+cosγcosα
Z 2 +
c
b
cosαcosβ
cosβ+cosγcosα
Z 3.
(11.5.4)
We could also proceed in this special case as follows. The argument is laid out
for the case in the diagram, withβandγacute angles and(Z 1 ,Z 2 ,Z 3 )positive in
orientation. The other cases can be treated similarly.
Now by 5.2.2 applied to[Z 10 ,Z 2 ,Z 3 ],|∠Z 11 Z 2 Z 4 |◦=|∠Z 10 Z 2 Z 3 |◦= 90 −|γ|◦so
that
|Z 11 ,Z 8 |
|Z 2 ,Z 8 |
=tan∠Z 11 Z 2 Z 8 =cotγ.
But|Z 2 ,Z 8 |=ccosβ and so|Z 11 ,Z 8 |=ccosβcotγ. ThusδF(Z 11 ,Z 2 ,Z 3 )=
1
2 accosβcotγand sinceδF(Z^1 ,Z^2 ,Z^3 )=
1
2 acsinβ,wehavethat
δF(Z 11 ,Z 2 ,Z 3 )
δF(Z 1 ,Z 2 ,Z 3 )
=cotβcotγ.
As similar results hold in the other two cases, we have by 11.4.2 that
Z 11 =cotβcotγZ 1 +cotγcotαZ 2 +cotαcotβZ 3.
That the sum of the coefficients is equal to 1 follows from the identity
tanα+tanβ+tanγ=tanαtanβtanγ