AB. Let this division of AD extend in DB to b, and draw the parallel bc. If the parts of AD and AB be again subdivided, the corresponding residue will evidently be diminished; and thus, at each successive subdivision, the terminating parallel bc must approximate perpetually to BC. Wherefore, by continuing this process of exhaustion, the divided lines Ab and Ae will approach the limits AB and AC, nearer than any finite or assignable interval. Consequent

E

& B

ly, from the preceding demonstration, AD: AB :: AE: AC.

And since AD:AB::AE: AC, it follows, by conversion (V. 11.), that AD: DB:: AE: EC, and again, by composition (V. 9.), that AB : DB : : AC : EC.

2. Let the two parallels DE and BC cut the diverging lines DB and EC, on opposite sides of A; the segments AB, AD have the same ratio with AC, AE,-or AB: AD:: AC: AE.

For, make AO equal to AD, AP to AE, and join OP. The triangles APO and

AED, having the sides AO, AP equal to AD, AE, and the contained vertical angle OAP equal to DAE, are equal (I. 3.), and consequently the angle AOP is equal to ADE; but these being alternate angles, the straight line OP (I. 25.) is

parallel to DE, and hence, from what was already demonstrated, AB: AO or AD: : AC: AP or AE.

And since AB : AD :: AC: AE, by conversion DB: DA:: EC: EA, and, by conversion, and inversion

DB: AB:: EC: AC.

3. Lastly, let more than two parallels, BC, DE, FH, and GI, intersect the diverging lines AB and AC; the seg ments DA, AF, FG, and GB, in DB, are proportional respectively to EA, AH, HI, and IC, the corresponding segments in EC.

For, from the second case, DA: AF:: EA: AH; and, from the first case, AF: FG :: AH: HI. But from the same

D

E

C

H

A

G B

case, AG: FG:: AI: HI, and AG: GB:: AI: IC; whence (V. 15.) FG: GB:: HI: IC.

Cor. 1. Hence the converse of the proposition is also true, or that straight lines which cut diverging lines proportionally are parallel; for it would otherwise follow, that a new division of the same line would not alter the relation among the segments, which is evidently absurd.

Cor. 2. Hence, if the segments of one diverging line be equal to those of another, the straight lines which join them are parallel.

PROP. II. THEOR.

Diverging lines are proportional to the corresponding segments into which they divide parallels.

Let two diverging lines AB and AC cut the parallels BC and DE; then AB : AD: : BC: DE.

For draw DF parallel to AC. And, by the last Proposition, the parallels AC and DF must cut the straight

A

E

Next, let more than two diverging lines AB, AF, and AC intersect the parallels BC and DE; the segments BF and FC have respective

ly to DG and GE the same ratio as AB has to AD.

From what has been already demonstrated, it appears, that AB: AD:: BF: DG, and also that AF AG :: FC: GE.

:

But by the last Proposition,

AB: AD::AF: AG; where

E

G

fore AB: AD:: FC: GE. The same mode of reasoning, it is obvious, might be extended to any number of sections. Whence AB: AD: : BF: DG::FC:GE.

Cor. 1. Hence straight lines which cut diverging lines equally, being parallel (VI. 1. cor. 1.), are themselves proportional to the segments intercepted from the vertex.

Cor. 2. Hence parallels are cut proportionally by diverging lines.

PROP. III. PROB.

To find a fourth proportional to three given straight lines.

Let A, B, and C be three straight lines, to which it is required to find a fourth proportional.

Draw the diverging lines DG and DH, make DE equal to A, DF to B, and DG to C, join EF, and through G draw (I. 26.) GH parallel to EF, and meeting DH in H; DH is a fourth proportional to the straight lines A, B, and C.

For the diverging lines DG and DH are cut proportion

A

B

H

F

D

E

ally by the parallels EF and GH (VI. 1.), or DE: DF:: DG: DH, that is, A: B:: C: DH.

Cor. If the mean terms B and C be equal, it is obvious, that DG will become equal to DF, and that DH will be found a third proportional to the two given terms A and B.

PROP. IV. PROB.

To cut a given straight line into segments which shall be proportional to those of a divided straight line.

Let AB be a straight line, which it is required to cut into segments proportional to those of a given divided straight line.

Draw the diverging line AC, and make AD, DE, and EC, equal respectively to the segments of the divided line, join CB, and draw EG and DF parallel to it (I. 26.) and meeting AB in F and G; AB is cut in

A

D

E

those points proportionally to the segments of AC.

B

For (VI. 2.) the parallels DF, EG, and CB must cut the diverging lines AB and AC proportionally (VI. 1.), or AF: FG:: AD: DE, and FG: GB:: DE: EC.

PROP. V. PROB,

To cut off the successive parts of a given straight line.

Let AB be a straight line from which it is required to cut off successively the half, the third, the fourth, the fifth, &c.

Through B draw the diverging straight line CBG continued both ways, take in it any point C, and make BD, DE, EF, FG, &c. each equal to BC, complete the parallelogram ABCI, and join ID, IE, IF, IG, &c. cutting AB in the points K, L, M, N, &c.; then is the segment AK the half of AB, AL the third, AM the fourth, and AN the fifth part, of the same given line.

For the segments of the straight line AB must be proportional to the segments of the parallels AI and BG, intercepted by the diverging lines ID,

E

IE, IF, IG, &c. Thus, AK: KB:: AI: BD; but, by construction, BC or AI=BD, whence (V. 4.) AK=KB, and therefore AK is the half of AB. Again, AL: LB:: AI: BE; and since BE-2AI, it follows, that LB=2AL, or AL is the third part of AB. In the same manner, AM: MB:: AI: BF; but BF=3AI, whence MB=3AM,