Theorem 2. A line can be divided, internally or externally, into segments having a given ratio, except that if it is divided externally the ratio cannot be unity.

Given ratio.

the line AB, and two lines S1, S2 having a given.

To prove that AB can be divided in the ratio s1: S2, except that in the case of external division s, cannot equal s2. Proof. 1. Suppose AM drawn making, with AB, an angle <180°; that AC be taken = s1, and CD=S2; that DB be drawn, and CP || DB.

2. Then AP: PB =S1 S2, as required.

Th. 1, cor.

1

3. In Fig. 2, if s1

then the relation of CP to AB? Hence show that the division is impossible in this case.

COROLLARY.

The point of internal division is unique; like

wise the point of external division.

From step 2, AB : PB = S1 + S2: S2, AB, S1 + S2, and S2, all being constants; but by three terms of a proportion the fourth is determined. (Def. 4th prop., cor. 1.)

NOTE. Instead of saying that the external division, if the ratio is unity, is impossible, it is often said that the point of division, P, is at infinity.

In the case of internal division, the ratios AP: PB and AC CD are evidently positive; but in the case of external division each ratio is evidently negative because PB and CD are negative.

EXERCISE. 397. If the three lines of a pencil O ABC cut off proportional segments on two other lines, these other lines are parallel.

Theorem 3. A line which divides two sides of a triangle proportionally is parallel to the third.

Given

the triangle ABC, and DE so drawn that AD: DC

Proof. 1. Suppose DE not II AB, but that DX II AB.

2. Then BX: XC = AD: DC.

3. But this is impossible,

ratio AD DC is unique.

Th. 1, cor. 1

the division of BC in the

Th. 2, cor.

4. .. DX must be identical with DE, and DE II AB.
The proof is the same for all of the figures.

EXERCISES. 398. In the above figures, if AD: DC = BE : EC = m :n, and if the line through A and E cuts the line through B and D at P, then prove that AP: PE = BP: PD = m+n: n.

399. If ex. 398 has been proved, show from it that the centroid of a triangle divides the medians in the ratio of 2: 1.

400. Prove th. 2 on the following figures:

S2

Si

B

401. From any point P in the common base AB of two triangles ABC and ABD, on the same side of AB, lines are drawn parallel to AC and AD, meeting BC, BD at F, G. Prove that FG || CD.

402. Investigate ex. 401 when P is on AB produced.

Theorem 4. If any angle of a triangle is bisected, internally or externally, by a line which also cuts the opposite side, then the opposite side is divided, internally or externally, respectively, in the ratio of the other sides of the triangle.

Given

A ABC, the bisector of C meeting AB at P.

To prove
that AP
Proof. 1. Let BE II PC,

2. Then EBC

3.

PB = AC:

BC.

meeting AC produced at E, in Fig. 1.

PCB =

ACP = CEB.

... BC=CE.

Given; I, th. 17, cor. 2

Why? Th. 1, cor. 1

Why?

6. The proof for Fig. 2 is the same if step 2 is

changed to

CBE =

BCP =

PCX =

BEC.

DEFINITION. When a line is divided internally and externally into segments having the same ratio, it is said to be divided harmonically.

If the internal and external points of division of AB, in th. 4, are P and P', then AB is divided harmonically by P and P'.

EXERCISES. 403. In ABC, suppose that a c, and the bisectors of the interior and exterior angles at C meet AB at P1, P2. Prove that if a circumference passes through P1, P2, and C, (1) P1 P2 is the diameter, (2) AC is a tangent.

404. The hypotenuse of a right-angled triangle is divided harmonically by any pair of lines through the vertex of the right angle, making equal angles with one of its arms.

Section 4.- A Pencil cut by Antiparallels or by a Circumference.

DEFINITIONS. If a pencil of two lines O-XY is cut by two parallel lines AB, MN, and if MN revolves, through a

B

B1

X

straight angle, about the bisector of XOY as an axis of symmetry, falling in the position A, B1, then AB and A,B1 are said to be antiparallel to each other.

OА and OẠ1 are called corresponding segments of the pencil, as are also OB and OB1. A and A1 are called corresponding points, as are also B and B1.

COROLLARY. If LA=ZA1, in the above figure, then AB and A1B1 are antiparallel to each other.

EXERCISES. 405. From P, a given point in the side AB of ▲ ABC, draw a line to AC produced so that it will be bisected by BC.

406. Investigate ex. 405 when P is on AB produced.

407. If the vertices of XYZ lie on the sides of ▲ abc so that x || a, y | b, z || c, then X, Y, Z bisect a, b, c. .

408. In th. 4, suppose B = ZA; also, suppose

B<ZA.

409. In any triangle the line joining the feet of the perpendiculars from any two vertices to the opposite sides is antiparallel to the third side.

Theorem 5. If a pencil of two lines is cut by two antiparallel lines, the corresponding segments form a proportion.

Given

the pencil OXY, cut by the antiparallels AB, A1B1, A and A, being corresponding points.

To prove

that OA OA1 = OB: OB1.

Proof. 1. Suppose MN the parallel to AB which, revolving,

COROLLARY. If two antiparallels cut a pencil of two lines, the product of the segments of one line equals the product of the segments of the other.

Why? What is meant by "product of two segments"?

EXERCISES. 410. In the figures on p. 152, AB A1B1 = OA: OA1 OB OB1. (Th. 1, cor. 2, etc.)

411. In the figures on p. 152, if A1 coincides with B, and if OB = b, OA = a, OB1 = b1, then b2 = abı.

412. If from the vertex of a right-angled triangle a perpendicular p is drawn cutting the hypotenuse c into two segments x, y, adjacent to sides a, b, respectively, then (1) a and p are antiparallels of the pencil b, c ; (2) a is a mean proportional between c and x; (3) p is a mean proportional between x and y ; (4) b2=cy, a2= cx, and .. a2+b2=c(x+y)=c2. (Thus a new proof is found for the Pythagorean proposition, and a subsequent theorem is proved at the same time.)

413. Show how th. 4 may be applied to bisect a line; to trisect it; to cut off 1/nth part of it.

414. ABCD is a quadrilateral.

Prove that if the bisectors of A, C meet on diagonal BD, then the bisectors of s B, D will meet on diag

onal AC.

415. Construct a triangle, having given the base, the vertical angle, and the ratio of the remaining sides. (Intersection of loci and th. 4.) 416. In ABC, CM is a median; BMC, CMA are bisected by lines meeting a and b in R and Q, respectively.

Prove that QR || AB.