Sec. 7.3 Formula for mid-line of an angle-support 107
Proof.Letlandmbe the perpendicular bisectors of[B,C]and[C,A], respectively.
Then if we hadl‖mwe would havel‖m,m⊥CAand sol⊥CAby 5.1.1; this would
yieldBC⊥l,CA⊥land soBC‖CAby 4.2.2(iv). This would make the pointsA,B,C
collinear and so give a contradiction.
Thuslmust meetmin a unique point,Dsay. Then by 4.1.1(iii)Dis equidistant
fromBandCas it is onl, and it is equidistant fromCandAas it is onm. Thus the
circle with centreDand length of radius|D,A|passes throughA,BandC.
Conversely, suppose that a circle passes throughA,BandC. Then by 4.1.1(ii) its
centre must be onland onmandsoitmustbeD. The length of radius then must be
|D,A|.
COROLLARY.Two distinct circles cannot have more than two points in common.7.3 Formula for mid-line of an angle-support ...............
7.3.1 .....................................
COMMENT. We now start to prepare the ground for our treatment of angles. Earlier
on we found that mid-points have a considerable role. Now we shall find that mid-
lines of angle-supports, dealt with in 3.6, have a prominent role as well. Given any
angle-support|BAC, if we take any numberk>0 there are unique pointsP 1 andP 2 on
[A,Band[A,Crespectively, such that|A,P 1 |=k,|A,P 2 |=k. ThusP 1 andP 2 are the
points of[A,Band[A,Con the circleC(A;k).Then|BAC=|P 1 AP 2 and it is far more
convenient to work with the latter form. We first prove a result which will enable us
to deal with the mid-lines of angle-supports by means of Cartesian coordinates.
With a frame of referenceF=([O,I,[O,J),let P 1 ,P 2 ∈C(O;1)be such that
P 1 ≡F(a 1 ,b 1 ),P 2 ≡F(a 2 ,b 2 ). Then the mid-line l of|P 1 OP 2 has equation
(b 1 +b 2 )x−(a 1 +a 2 )y= 0when P 1 and P 2 are not diametrically opposite, and equation a 1 x+b 1 y= 0 when they
are.
Proof.WhenP 1 andP 2 are not diamet-
rically opposite, their mid-pointMis
notOand we havel=OM.AsMhas
coordinates
( 1
2 (a^1 +a^2 ),1
2 (b^1 +b^2 ))
,
the line OM has equation
(b 1 +b 2 )x−(a 1 +a 2 )y=0. WhenP 1
is diametrically opposite toP 2 ,lis the
line throughOwhich is perpendicular
toOPand this has the given equation.
P 1
P 2
P 3
M
O
Q
Figure 7.3.