Sidebilder
PDF
ePub

Let the three points a, b, c be taken in the sides of ABC, so that

A

aC.A.cB=aB.C.CA; then will abc be a straight line.

For if not, let abd' be a straight line, then (Prop. 8)

a C.b1.cB-aB.6C.CA. But (Hyp.),

aC.bA.cB=aB.6C.CA,

c'B C'A whence, by division

; but this can only be true cB

CA when d coincides with c; hence abc is a straight line.

B

[ocr errors]

Proposition 10. Theorem.-The three lines drawn from any point in the plane of a triangle to the three angles, and cutting the opposite sides ( produced if necessary), will cut them so that the product of three non-adjacent segments is equal to the product of three others.

Let the three lines, A , BOB, COC, be drawn through the angles of ABC and any point, O, in its plane, then aB.bc.cA= aC.bA.cB.

Because the triangle A Ca is cut by the transversal Bb, (Prop. 8), Ba.04.6C-BC. Oa .bA. Also because the triangle ABa is cut by the transversal Cc,

CB. Oa.cĄ = Ca. OA.cB. Multiplying these together, we have

aB.6C.CA = aC.6A.cB.

A

B

a

Proposition 11. Theorem.-If three points be taken on the sides of a triangle, so that the product of three non-adjacent segments is equal to the product of the other three, the three straight lines joining these points with the opposite angles will pass through the same point.

The proof is similar to (Prop. 9).

From the last theorem the following propositions may be proved :

1. The three lines drawn from the angles of a triangle to the middle points of the opposite sides intersect in the same point.

2. The three lines drawn bisecting the angles of a triangle intersect in the same point.

3. The three perpendiculars drawn from the angles of a tri. angle to the opposite sides intersect in the same point.

Proposition 12. Theorem.-In a complete quadrilateral each diagonal is divided harmonically by the other two.

Let the diagonal EF be cut by the diagonals AC, BD in P and Q; then will

PE : PF :: QE : QF. Because the transversal QDB cuts the triangle AEF, we have (Prop. 8)

QE. DF.BA = QF.DA.BE.

Because the straight lines CA, CF, CE cut the opposite sides of the triangle AEF, we have (Prop. 10)

PE.DF. BA= PF. DA. BE. Dividing one of these equations

PE PF by the other,

QE

A

В.

E

P

whence PE : PF :: QE : QF; hence EF is divided harmonically in P and Q.

The same may be proved of either of the diagonals AC, BD.

Proposition 13. Theorem.-The middle points of the three diagonals of a complete quadrilateral are in the same line.

F

D

A

B

Let L, M, N be the three middle points of the diagonals AC, BD, EF.

LMN is a straight line.

Bisect the sides of the triangle BCE in G, H and K. The line HK will be parallel to AE, and will meet AC in L; similarly, GH will pass through N, and GK through M.

Considering FDA as a transversal of the triangle EBC (Prop. 8), gives

EA.BF. CD= ED, BA. CF.

G

Taking the half of each of these lines,

HL.GN. MK = GM. KL.HN.

Hence, LMN may be considered a transversal of the triangle GHK, and is a straight line (Prop. 9).

SECTION V.

ANHARMONIC AND HARMONIC RATIO.

DEFINITIONS.

[ocr errors]

D

B
1

1. The anharmonic ratio of four points is the quotient obtained by dividing the ratio of the distances of the first two from the third, by the ratio of the distances of the first two from the fourth.

Thus, the anharmonic ratio of the four points A, B, C, D is

AC AD

BC BD' and is denoted for brevity by [ABCD).

BA BD The anharmonic ratio (BCAD] would be

CA CD 2. If the anharmonic ratio of four points be equal to unity, they are said to be in harmonic ratio. AC AD

AC AD Thus, if

= 1, BC BD

BC BDA, B, C, D are harmonic points, which agrees with Def. 5,

or

page 245.

3. A pencil is a system of straight lines diverging from a point.

4. A ray is one of these diverging lines.

5. The vertex of the pencil is the point from which they diverge.

Proposition 14. Theorem.-The anharmonic ratio of a system of points is not changed by interchanging two of the points, provided the other two points be also interchanged.

For

[ABCD]

AC AD AC.BD

- ;
BC BD BC.AD
BD BC AC.BD
AD AC BC.AD

also,

[BADC]

Hence, also, [ABCD] =[BADC] =[CDAB]=[DCBA).

Scholium. There are twenty-four ways in which these four letters may be arranged; hence there are six different anharmonic ratios which may be formed from them.

Proposition 15. Theorem.-If a pencil of four rays be cut by any two transversals, the anharmonic ratio of the four cutting points of one transversal is equal to the anharmonic ratio of the four corresponding points of the second.

Let OM, ON, OP, OQ be four rays, and ABCD, A'B'C'D any two transversals; then will [ABCD] =[A'B'C'D'].

B/ cla\D
From B and B draw Ba,
B'b parallel to AO, cutting
OC" in c and d.

From similar triangles CAO,
CBc, and also DAO, DBa,

A

D

BI

A

M

CA : CB :: OA : cB,

and

DA : DB :: OA : aB.

« ForrigeFortsett »