Geometry with Trigonometry

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