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


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 is to A, DF to B, and DG to CH- I-II C, join EF, and through G

draw (I. 23.) GH parallel to I.
EF and meeting DH in H;
DH is a fourth proportional
to the straight lines A, B, and IX T. G.

C. 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 : Br: : 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.


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

straight line. M

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. 23.) and meeting AB in G and F; AB is cut in those points proportionally to the segments of AC.

For 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.


To cutoff the successive parts of a given straight line.

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

On AB describe (I. 23.) the rhomboid ABCD, and through E, the intersection of its diagonals AC and BD, draw EF parallel to AD, join DF, and through G, where it cuts AC, draw GH likewise parallel to AD, again join DH and draw the parallel IK, and so repeat the operation: Then will AF be the half of AB, AH the third, AK the fourth, and AM the fifth part of it.

The triangles AED and CEB are equal (I.20), since



they have (I.23.) the angles DAE and ADE equal to BCE and CBE, and the interjacent sides AD and CB (I. 26.) likewise equal; and therefore DE=EB. But AD and EF being parallel, DE: EB:: AF : FB(VI.1.); whence (V.4.) AF=FB, or AF is the half of AB. And AD and EF being intercepted parallels, AD : EF; ; AB : BF (VI. 2.); consequently, since AB is double of BF, AD is likewise double of EF (V. 5.).—Again, the I} C diverging lines AGE and DGF are proportional to the intercepted parallels AD and EF (VI. 2.), or AD : EF : : AG : GE ; and GH being parallel to EF, AG : GE : : AH : HF (VI. 1.), whence AD : EF: : AH : HF; but AD was shown to be double of EF, wherefore AH is double of HF (V. 5.), or AH is two-thirds of AF, or of the half of AB, and is consequently the third part of the whole AB. Now, since AF: HF: ; AD : GH, (VI.2) and AFis triple of HF, it is evident that AD is triple of GH; but AD : GH :: AI : IG :: AK : KH, and, AD being triple of GH, AK must also be triple of KH; or AK is three-fourths of AH, which was proved to be the third of AB, whence the segment AK is the fourth part of the whole line A.B. By a like process, it is shown that AM is the fifth part of AB. Cor. This construction likewise exhibits other portions of the line AB. For, since AF is the half, and AH the third, their difference FH must be the sixth part. Again, AH and AK being the third and fourth parts, the interval HK is the twelfth. In like manner, it is shown that KM is the twentieth part of AB. -

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To divide a straight line harmonically, and in a given ratio.

Let AB be a straight line, which it is required to cut harmonically, in the ratio of K to L.

Through A draw the diverging line AC, and produce it both ways till AC and AD be each equal to K, make

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CB, draw EF parallel | I. |
to AB, and FG paral- |
lel to CA, and join - T

DF cutting AB in H; the straight line AB is divided harmonically in the points H and G, such that K: L:: AB : BG : : AH : HG.

For the parallels AC TX and GF, being intercepted by the diverging lines AB and CB, AC: GF :: AB: BG (VI. 2.). Again, the diverging lines AG and DF are cut by the parallels AD and FG, whence (VI. 2.) AD or AC: GF : : AH : HG. Wherefore, AB : BG : : AH : HG; and each of these ratios is the same as that of AC or AD to GF, or that of K to L.

Cor. Hence AG is divided, internally in H and externally in B, in the same ratio. In like manner, BH is divided proportionally, by an external and internal section in A and G; for AB : BG : : AH : HG, and alternately AB : AH : ; BG : HG,



If a straight line be divided internally and externally in the same ratio, half the line is a mean proportional between the distances of the middle from the two points of unequal section.

Let the straight line AB be divided in the same ratio, internally and externally in C and D, and also be bisected in E; the half EB is a mean proportional between EC and . ED, or EC: EB :: EB : ED. For since AC : CB :: AD : DB, by mixing and inversion AC–CB: AC+CB: ; AD–DB : AD+DB, that is, 2EC : AB :: AB : 2ED, and, halving all the terms of the analogy, (V. 3.) EC: EB :: EB : ED. Cor. Hence if a straight line be cut internally and exter- o mally in the same ratio, the square of the interval between the points of section is equivalent to the difference between the rectangles under the internal and external segments. For (II. 17.) AD.DB=ED"—EB”, and AC.CB=EB*— EC*; consequently AD.DB–AC.CB = ED*—2EB*-iEC*, or (V. 6. and VI, 7.) ED*—2ED.EC+EC", which (II. 16.) is the square of ED–EC or of CD.

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If diverging lines divide a straight line harmo- mically, they will cut every intercepted straight line also in harmonic proportion.

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