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242°. Theorem.---The mean centre of the vertices of a triangle with multiples proportional to the opposite sides is the centre of the incircle.

Proof.-Take L along one of the sides, as BC, and let p be the from A. Then Σ(a.AL)=a.p

and

.. (241°)

(a). OL=(a+b+c). OL,

OL=

ар

B

Р

C

ap__=r; (153°, Ex. 1)

a+b+c S

i.e., the mean centre is at the distance r from each side, and is the centre of the incircle.

Cor. I. If one of the multiples, as a, be taken negative,

OL=

-ap
- a+b+c

-^=-r';

s-a

(153°, Ex. 2)

i.e., the mean centre is beyond L, and is at the distance r' from each side, or it is the centre of the excircle to the side a.

Cor. 2. If any line be drawn through the centre of the incircle of a triangle, and a, ß, y be the perpendiculars from the vertices upon it, aa+bB+cy=0,

and if the line passes through the centre of an excircle, that on the side a for example, aa=bB+cy.

EXERCISES.

I. If a line so moves that the sum of fixed multiples of the perpendiculars upon it from any number of points is constant, the line envelopes a circle whose radius is

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2. The mean centre of the vertices of a triangle, for equal

multiples, is the centroid.

3. The mean centre of the vertices of any regular polygon, for equal multiples, is the centre of its circumcircle.

N

243°. Theorem.-If O be the mean centre of a system of points for a system of multiples, and P any independent point in the plane,

and

.B

A

(a. AP2)=2(a. AO2)+2(a). OP2.

Proof-Let O be the mean centre, P the independent point, and A any point of the A system. Let L pass through O and be perpendicular to OP, and let AA' be perpendicular to OP. Then

AP2=AO2+OP2 – 2OP. OA',

a. AP2=a. AO2+a. OP2 - 2OP. a. OA'.

Similarly 6. BP2=b. BO2+b. OP2 - 2OP. 6. OB',

But

(a. AP2)=2(a. AO2)+2(a). OP2-20P. Z(a. OA'). (a. OA')=2(a. AL)=0,

(a. AP2)=2(a. AO2) + 2(a). OP2.

(241°)

q.e.d.

Cor. In any regular polygon of n sides th the sum of the

n

squares on the joins of any point with the vertices is greater than the square on the join of the point with the mean centre of the polygon by the square on the circumradius.

For making the multiples all unity,

(AP)=nr2+nOP2,

1Σ(AP2)=OP2+r2.

n

Ex. Let a, b, c be the sides of a triangle, and α, β, γ the joins of the vertices with the centroid. Then (242°, Ex. 2)

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Ex. If ABCDEFGH be the vertices of a regular octagon taken in order, AC2+AD2+AE2+AF2 + AG2= 2(6+ √2)r2.

244°. Let O be the centre of the incircle of the ABC and let P coincide with A, B, and C in succession.

Ist.

2nd.

bc2+cb2=2(a. AO2) + 2(a)AO2,

ac2 +ca2=2(a. AO2)+2(a)BO2,

3rd. ab2+ba2 =2(a. AO2)+2(a)CO2.

Now, multiply the 1st by a, the 2nd by b, the 3rd by c, and add, and we obtain, after dividing by (a+b+c),

Σ(a. AO2)=abc.

Cor. I. For any triangle, with O as the centre of the incircle, the relation Σ(a. A P2)=2(a. AO2)+2(a)OP2

becomes

Z(a. AP2)=abc+2s. OP2,

and, if O be the centre of an excircle on side a, for example, Z(a. AP2)=abc+2(sa)OP2,

where a denotes that a alone is negative.

Cor. 2. Let P be taken at the circumcentre, and let D be the distance between the circumcentre and the centre of the incircle. Then

...

But

and

AP BP CP=R.

2sR2=abc+2sD2.

abc=4AR,
s=4

r

D2 R2-2Rr.

(204°, Cor.)

(153°, Ex. 1)

Cor. 3. If D1 be the distance between the circumcentre and the centre of an excircle to the side a, we obtain in a similar

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245°. Ex. To find the product OA. OB. OC, where O is the centre of the incircle.

Let P coincide with A. Then (244°)

bc2+b2c=abc+2s. AO3,
AO2 = bc(s -- a)

S

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246°. If (a. AP2) becomes constant, k, we have

k=2(a. AO2)+2(a)OP2,

and (a. AO2) being independent of the position of P, and therefore constant for variations of P, OP is also constant, and P describes a circle whose radius is

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.. If a point so moves that the sum of the squares of its joins with any number of fixed points, each multiplied by a given quantity, is constant, the point describes a circle whose centre is the mean point of the system for the given multiples.

EXERCISES.

1. If O', O′′, O′′ be the centres of the escribed circles,

2.

3.

AO'. BO". CO""=4Rs2.

AO'. BO'. CO'=4Rr12.

s.OL=(s− a)O'L+(s−b)O′′L+(s−c)O′′L,

where L is any line whatever.

4. If P be any point,

5.

s. OP2=(-a)O'P2 + (s-b)O"P2+(s-c)O""P2 - 2abc. (sa). O'O2+(s-b). O′′O2 + (s-c)O""O2=2abc.

6. If D is the distance between the circumcentre and the

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SECTION III

OF COLLINEARITY AND CONCURRENCE.

247°. Def. 1.—Three or more points in line are collinear, and three or more lines meeting in a point are concurrent.

Def. 2.—A tetragram or general quadrangle is the figure formed by four lines no three of which are concurrent, and no two of which are parallel.

Thus L, M, N, K form a tetra

gram. A, B, C, D, E, F are its six vertices. AC, BD are its internal

N

M

C

K

diagonals, and EF is its external diagonal.

248°. The following are promiscuous examples of collinearity and concurrence.

Ex. 1. AC is a O, and P is any point. Through P, GH is drawn to BC, and EF || to AB.

K

N

G

A

B

P

E

M

S

F

The diagonals EG, HF, and DB of the three

AP, PC, and AC are concur

rent.

EG and HF meet in some D

H

с

point O; join BO and complete the OKDL, and make

the extensions as in the figure.

We are to prove that D, B, O are collinear.

Proof-KG÷□GM, and FL=OFN, (145°)

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