Sec. 11.5 Vector methods in geometry 205

for the angles of a triangle. For, using the notation of 10.8.1,|α|◦+|β|◦+|γ|◦=180,

and soαF+βF+γF= (^180) Fso that
−tanγF=tan( (^180) F−γF)=tan(αF+βF)

tanαF+tanβF
1 −tanαFtanβF

,

whence the result follows by multiplying across and rearranging. This formula fails

in the case of a right-angled triangle.

From our two methods we have two formulae forZ 11 , but we further note that

cos(αF+γF)=cos( (^180) F−βF),
cosαFcosγF−sinαFsinγF=−cosβF,
cosαcosγ+cosβ=sinαsinγ.
On using this with the sine rule, the two formulae for the orthocentre are recon-
ciled.

11.5.7 Incentre of a triangle..........................

LetZ 1 ,Z 2 ,Z 3 be non-collinear points. By 5.5.1 the mid-line of|Z 2 Z 1 Z 3 will meet

[Z 2 ,Z 3 ]in a pointZ 12 whereZ 12 =( 1 −r)Z 2 +rZ 3 ,and

r

1 −r

=

c

b

.

By similar arguments the mid-line of|Z 3 Z 2 Z 1 will meet[Z 3 ,Z 1 ]in a pointZ 13 where

Z 13 =( 1 −s)Z 3 +sZ 1 ,and

s

1 −s

=

a

c

,

and the mid-line of|Z 1 Z 3 Z 2 will meet[Z 1 ,Z 2 ]in a pointZ 14 whereZ 14 =( 1 −t)Z 1 +
tZ 2 ,and
t
1 −t

=

b

a

.

The product of these three ratios is clearly equal to 1 so (11.5.1) is satisfied. By the

cross-bar theorem,[Z 2 ,Z 13 will meet[Z 3 ,Z 14 ]in a pointZ 15 and soZ 2 Z 13 ,Z 3 Z 14 meet

inZ 15. It follows thatZ 1 Z 12 also passes through the pointZ 15.

Thus the mid-lines of the

angle-supports |Z 2 Z 1 Z 3 ,

|Z 3 Z 2 Z 1 ,|Z 1 Z 3 Z 2 for a triangle

[Z 1 ,Z 2 ,Z 3 ]are concurrent.The

perpendicular distances from

this pointZ 15 to the side-lines

of the triangle are equal by

Ex.4.4, so the circle withZ 15

as centre and length of radius

these common perpendicular

distances will pass through the

feet of these perpendiculars.

Z 1

Z 2

Z 3

Z 15

Figure 11.11. Incentre of triangle.