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C19.

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Book VI. ratio of AB to FGC, or of EB to LG; but EBC is likewise fie

milar to LGH, and are to one another in the duplicate ratio of EB to LG, or of EC to LH; and, for the same reason, ECD is to LHK in the duplicate ratio of CE to LH; therefore the triangles in the polygon ABCDE are equal in number to the triangles in the polygon FGHKL, and similar to one another; therefore, because the triangle ABE is to the triangle FGL in the duplicate ratio of BE to GL; and the triangle EBC to the triangle LGH, in the duplicate ratio of BE to GL; therefore the triangles. ABE, EBC, are to FGL, LGH, as EBC is to LGH d. For the same reason, EBC, ECD, are to LGH, LHK, as EBC is to LGH: Therefore all the antecedents ABE, EBC, ECD, are to all the consequents FGL, LHG, LHK, as ABE is to FGL"; that is, in the duplicate ratio of AB to FG'; and polygon to polygon in the duplicate ratio of one homologous fide to another. Wherefore, &c.

COR. Hence, if three right lines are proportional, the polygon described on the first is to the similar polygon described on the second as the first is to the third; for, if X be taken a third proportional to any two right lines, AB, FG ; then AB is to X in a duplicate ratio of AB to FG ; that is, any fimilar figures described on AB, FG, are to one another in the duplicate ratio of AB to FG.

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IGURES that are similar to the same right lined figures are also similar to one another.

a def. 1.

Let each of the right lined figures A, B, be similar to the right lined figure C; then the right lined figure A will be fimilar to the right lined figure B.

For, because the right lined figure A is similar to C, it is equiangular to it a; and the sides about the equal angles proportional. For the same reason, B is equiangular to C, and the fides about the equal angles proportional; therefore each of the

figures A, B, are equiangular to C; and therefore equiangular b ax. 1. 1. to one another b, and the lides about the equal angles proportion

al'; wherefore A is fimilar to B. Wherefore, &c.

CII. 5

PRO P. XXII. THE O R.

Book VI.

IF F four right lines are proportional, the right lined figures fimi

lar, and similarly described upon them, are proportional; and, if similar right lined figures similarly described upon right lines be proportional, the right lines fall also be proportional.

Let four right lines AB, CD, EF, GH, be proportional, viz. as AB is to CD fo is EF to GH; on AB, CD, let the fimilar figures KAB, LCD, be similarly described a; and upon a 18. EF, GH, let MF, NH, be described fimilar to one another; then KAB will be to LCD as MF is to NH.

For, to AB, CD, take X a third proportionaló, and O, a third b 11. proportional to EF, GH. Now, because AB is to CD as EF is to GH, and CD is to X as GH is to O, then AB is to X as EF is to Obut, as AB is to X, so is the right lined figure KAB c 22. 5. to the similar figure LCD d; and, as EF is to O, fo is MF to d cor. 200 NH 4; therefore, as the right lined figure KAB is to the similar figure LCD, so is the right lined figure MF to the similar figure NH"; and, if KAB is to LCD as MF is to NH, then < 11. si AB is to CD as EF is to GH.

For, if not, let AB be to CD as EF is to PRF; upon PR de- f 12. scribe a figure SR similar to MF or NH; then KAB is to LCD as MF is to SR, and as MF is to NH ; therefore SR,NH, have the same proportion to MF; therefore SR is equal to NHS, and 5 9. s. also similar to it; therefore PR is equal to GH; therefore AB is to CD as EF is to GH. Wherefore, &c.

PRO P. XXIII. T H E O R.

one another that is compounded of their fides.

Let AC, CF, be equiangular parallelograms, having the angle BCD equal to the angle ECG; then the parallelogram AC, to the parallelogram CF, is in the proportion compounded of their fides, viz. of BC to CG, and DC to CE; for, place BC in a right line with CG, and DC in a right line with CE ", and compleat the parallelogram DG; then, as BC is a 14. I. to CG, so let K be to L; and, as DC is to CE, so let L be to Mb; but the ratio of K to M is compounded of the ra-b 12. tios of K to L, and L to M°; therefore the ratio of K to M is c def. Se that compounded of BC to CG, and DC to CE ; but BC is to CG as AC is to DG d; and DC is to CE as DG is to CF d; but BC is to CG as K to L, and DC to CE as L to M; therefore

AC

di.

Book VI. AC is to CF as K to Me that is, as BC to CG, and DC to

CE. Wherefore, &c.

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PRO P. XXIV. THEO R.

IN every parallelogram the parallelograms that are about the dia

ameter are fimilar to the whole, and also to one another.

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In the parallelogram ABCD the parallelograms GE, KH, are similar to the whole ABCD, and likewise to one another. For, in the triangles ADC, AGF, the angle AGF is equal to the angle ADC, and the angle AFG to ACD“; and the angle GAF common to both; therefore the triangles AGF, ADC, are equiangular. For the fame reason, the triangles AEF, ACB, are equiangular, and the sides about the equiangles pro. portional b. Again, in the triangles AGF, FKC, the angle GAF is equal to the angle KFG”,

and AGF equal to FKČ; for each are equal to the angle ADC”, and AFG to FCK; therefore the triangles AGF, FKC, are equiangular, and the fides about the equal angles proportional b. For the same reafon, AEF is equiangular to FCH; and the sides about the equal angles proportional. Then, because the two angles KFC, HFC, are equal to the two angles GAF, EAF; that is, the whole angle KFH equal to the whole angle GAE, and the angle C common to both; and the angles at K, H, equal to the angles at D, B, each to each; the parallelogram KH is equiangular to DB; for the same reason GE is equiangular to DB; therefore KH is equiangular to GE; and the sides about the equal angles proportional : For, because the angles GAE, KFH, are equal

, GA is to AE as KF is to FH; but KF is to FH as DA is to AB, for each are proportional to AF, FC. For the fame reafon, the sides about the other angles are likewise proportional; therefore the parallelogram DB is fimilar to KH; but GE is likewise similar to KH; therefore GE is fimilar to DB. Where.

CII. S.

fore, &c.

PRO P. XXV. PRO B.

To describe a figure
similar to a given right lined

figure, and e. qual to another given right lined figure.

LetABC and D, be two given right lined figures; it is required to describe a right lined figure similar to ABC, and equal to D.

On

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e 18.

On the fide BC, of the given figure ABC, make a parallelo- Book VI, gram, BE, equal to it a; and on the side CE make the parallelogram CM equal to the right lined figure D b; and the angle a 42.i. FCE equal to the angle CBL b; then BC, CF, as alfo LE, EM, 44. I.

& 29. 1. will be right lines Find GH a mean sproportional to BC, CFd; and on GH describe the right lined figure KGH similar 14. 1. and alike fituate to ABCC; then, because BC is to GH as GH £ 13. is to CF; and, as BC is to CF, so is the right lined figure ABC to the right lined figure KGHf; but, as BC is to CF, fo is the parallelogram BE to EF8; therefore, as the right lined figure f Cor. 20, ABC is to the right lined figure KGH, so is the parallelogram 8 1. BE to the parallelogram EFh; altern. as ABC is to BE, so is KGH to EF: But the right lined figure ABC is equal to the parallelogram BE; therefore the right lined figure KGH is equal to the parallelogram FE'; but FE is equal to the right lined figure D; therefore the right lined figure KGH is equal to D: Which was required.

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i 14. S.

PRO P. XXVI. THEO R.

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F, in a parallelogram, be constitute another parallelogram fimi . mon with it, they shall be about the same diameter.

Let the parallelogram AF be constitute in the parallelogram
ABCD, similar to it, and alike situate, having the angle DAB
common to both; then the parallelograms ABCD and AF are a.
bout the fame diameter AC.

For, if not, let AHC be the diameter of the parallelogram
BD, and produce GF to H; draw HK parallel to AD or BC;
then, because the parallelograms ABCD, KG, are about the
same diameter, they will be similar to one another à; and DA to
AB as GA to KA'b; but, because the parallelograms ABCD, A 34.

b Def. I. GE, are likewife fimilar", DA is to AB as GA is to AE

с Нур. therefore, as GA is to AE fo is GA' to AK ; therefore AE is equal to AKę, the greater to the less, which is impossible ; therefore the parallelograms AH, ABCD, are not about the fame diameter AHC; therefore no other but AF can be about the fame diameter with ABCD. Wherefore, &c.

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d II. S.

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Book VI.

PRO P. XXVII. T H E O R.

F all parallelograms applied to the same right line, and want,

ing in figure by parallelograms similar and alike ftuate to that described on half the line, the greatest is that which is applied to the half line, and similar to the defect.

Let AB be a right line bifected in the point C; and let the parallelogram AD be applied to the right line AB, wanting in figure by the parallelogram CE, similar and alike situate to that described on half the line AB; then AD is greater than a parallelogram applied to any other part of the right line AB, wanting in figure by a parallelogram similar and alike fituate to CE. For, let the parallelogram AF be applied to the right line AB, wanting in figure by the parallelogram HK, similar and alike situate to ČE; then the parallelogram AD is greater than AF. For, because the parallelogram CE is similar to HK, they will stand about the fame diameter. Let DB, that diameter,' be drawn, and the figure described ; then the parallelograms CF, FE, are equal b; add HK, which is common to both; then the whole CH is equal to the whole KE, but CH is equal to GCC; add CF, which is common; then the whole AF is equal to the gnomon EKN but the parallelogram CE is greater than the gnomon EKN ; therefore CE, that is, AD, is greater than AF. Wherefore, &c.

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PON a given right line to apply a parallelogram equal to a

given right lined figure, and deficient by a parallelogram fimilar to a given parallelogram; but the right lined figure to which the parallelogram is to be made equal, must not be greater than that described on half the line, as the defect must be similar.

It is required, upon the given right line AB, to apply a parale lelogram equal to the right lined figure C, and deficient by a parallelogram similar to D; and the right lined figure C not greater than the parallelogram described on half the line AB, which is similar to D. For, bisect AB in E, and on EB describe a parallelogram EF similar and alike situate to D4, and compleat the parallelogram AG,

Now, AG is either equal or greater than C; if equal, what was required is done. If not, make the parallelogram KLMN

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