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PROP. I. THEOR.

Parallels cut diverging lines proportionally.

The parallels DE and BC cut the diverging lines, AB and AC into proportional segments.

Those parallels may lie on the same side of the vertex, or on opposite sides; and they may consist of two, or of more straight lines.

1. Let the two parallels DE and BC intersect the diverging lines AB and AC, on the same side of the vertex A; then are AB and AC cut proportionally, in the points D and E,-or AD: AB:: AE: AC.

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For if AD be commensurable with AB, find (V. 26.) their common measure M, which repeat from the vertex A to B, and, from the corresponding points of section in AD and AB, draw (I. 23.) the parallels FI, GK, and HL. It is evident, from Book I. Prop. 36. that these parallels will also divide the straight lines AE and AC equally. Wherefore the measure M, or AF the submultiple of AD, is contained in AB, as often as AI, the like submultiple of AE, is contained in AC; consequently (V. def. 10.) the ratio of AD to AB is the same with that of AE to AC.

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But if the segments AD and AB be incommensurable, they may still be expressed numerically, to any required degree of precision. For AD being divided (I. 36.) into equal sections, these parts, continued towards B, will, together with some residuary portion, compose the whole of

AB. Let this division of AD extend through DB as far as 6, and draw the parallel bc. Let the parts of AD and AB be again subdivided, and the corresponding residue will evidently be diminished; consequently, at each successive subdivision, the terminating parallel bc will approximate perpetually to BC. Wherefore, by continuing this process of exhaustion, the divided lines Ab and Ac will approach their limits AB and AC, nearer than any finite

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or assignable interval. Consequently, 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

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being alternate angles, the straight line OP (I. 22.) is parallel to DE or BC, and hence, from what was already demonstrated, AB: AO or AD:: AC: AP or AE.

And since AB AD; : AC: AE, by composition BD :

AD: CE AE, and, by conversion, BD: AB:: CE: AC.

3. Lastly, let more than two parallels, BC, DE, FH, and GI, intersect the diverging lines AB and AC; the segments 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, AD: AF: AE: AH; and, from the first case, AF: FG:: AH: HI. But from the same

case, AG: FG :: AI: HI, and AG GB :: AI : IC;

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whence (V. 15.) FG : GB :: HI: IC.

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Cor. 1. Hence the converse of the proposition is also true, or straight lines which cut diverging lines pros portionally 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 lines AB and BC proportionally, or AB: AD:: BC: CF. But CF is equal (I. 26.)

to the opposite side DE of the parallelogram DECF; and consequently AB: AD::" BC: DE.

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

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AF: AG; wherefore AB : AD: : FC: GE. The same mode of reasoning, it is obvious, might be extended to any number of sections.

BF: DG:: FC: GE.

Whence AB: AD: :

Cor. 1. Hence the straight lines which cut diverging lines equally, being parallel (VI. 1. cor. 2.), are themselves proportional to the segments intercepted from the vertex.

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

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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. 23.) GH parallel to EF and meeting DH in H;

DH is a fourth proportional

to the straight lines A, B, and

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For the diverging lines DG and DH are cut proportionally 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.

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