C 19.

Book VI. ratio of AB to FG, or of EB to LG; but EBC is likewife fimilar to LGH, and are to one another in the duplicate ratio of EB to LG, or of EC to LH; and, for the fame 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 fimilar 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 4. For the fame reason, EBC, ECD, are to LGH, LHK, as EBC is to LGH: Therefore all the antecedents ABE, EBC, ECD, are to all the confequents 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.

€ 12. 5.

COR. Hence, if three right lines are proportional, the polygon defcribed on the firft is to the fimilar 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 defcribed on AB, FG, are to one another in the duplicate ratio of AB to FG.

a def. 1.

F

IGURES that are fimilar to the fame right lined figures are alfo fimilar to one another.

Let each of the right lined figures A, B, be fimilar 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 fimilar to C, it is equiangular to it; and the fides about the equal angles proportional. For the fame 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 bax. 1. 1. to one another, and the fides about the equal angles proportional'; wherefore A is fimilar to B. Wherefore, &c.

C II. 5.

PRO P. XXII. THE OR.

I
F four right lines are proportional, the right lined figures fimi-
lar, and fimilarly defcribed upon them, are proportional; and, if
fimilar right lined figures fimilarly defcribed upon right lines be
proportional, the right lines fhall 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 fimilarly described ; and upon EF, GH, let MF, NH, be defcribed fimilar to one another; then KAB will be to LCD as MF is to NH.

Book VI.

a 18.

For, to AB, CD, take X a third proportional, and O, a third b 11. proportional to EF, GH. Now, becaufe 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 0; but, as AB is to X, fo is the right lined figure KAB c 22. 5. to the fimilar figure LCD d; and, as EF is to O, fo is MF to d cor. 20 NH4; therefore, as the right lined figure KAB is to the similar figure LCD, fo is the right lined figure MF to the fimilar figure NH; and, if KAB is to LCD as MF is to NH, then e 11. 5 AB is to CD as EF is to GH.

For, if not, let AB be to CD as EF is to PR f; upon PR de- f 12fcribe a figure SR fimilar 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 fame proportion to MF; therefore SR is equal to NH, and 5 9. 5. alfo fimilar 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.

Eone the proportion to

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 AČ, 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. In to CG, fo let K be to L; and, as DC is to CE, fo let L be to M ; but the ratio of K to M is compounded of the ra-b 12. tios of K to L, and L to Mc; therefore the ratio of K to M is c def. 5, that compounded of BC to CG, and DC to CE; but BC is to CG as AC is to DG; 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

BOOK VI. AC is to CF as K to M; that is, as BC to CG, and DC to CE. Wherefore, &c.

e 22. S.

a 29. I.

b 4.

CII. S.

PRO P. XXIV. THEOR.

IN every parallelogram the parallelograms that are about the diameter are fimilar to the whole, and also to one another.

In the parallelogram ABCD the parallelograms GE, KH, are fimilar to the whole ABCD, and likewife 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 reafon, the triangles AEF, ACB, are equiangular, and the fides about the equiangles proportional. Again, in the triangles AGF, FKC, the angle GAF is equal to the angle KFG, and AGF equal to FKC; for each are equal to the angle ADCa, and AFG to FCK; therefore the triangles AGF, FKC, are equiangular, and the fides about the equal angles proportional. For the fame reafon, AEF is equiangular to FCH; and the fides 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 fame reason GE is equiangular to DB; therefore KH is equiangular to GE; and the fides 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 fides about the other angles are likewife proportional; therefore the parallelogram DB is fimilar to KH; but GE is likewife fimilar to KH; therefore GE is fimilar to DB. Wherefore, &c.

PRO P. XXV. PRO B.

To defcribe a figure fimilar to a given right lined figure, and equal to another given right lined figure.

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

d 13.

e

On the fide BC, of the given figure ABC, make a parallelo- Book VI, gram, BE, equal to it; and on the fide CE make the parallelogram CM equal to the right lined figure D b; and the angle a 42.1. FCE equal to the angle CBL ; then BC, CF, as alfo LE, EM, b 44. I. will be right lines. Find GH a mean proportional to BC, & 29. 1. CFd; and on GH defcribe the right lined figure KGH fimilar 14. 3. and alike fituate to ABC; then, because BC is to GH as GH is to CF; and, as BC is to CF, fo is the right lined figure ABC € 18. to the right lined figure KGH f; 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, fo is the parallelogram BE to the parallelogram EFh; altern. as ABC is to BE, fo 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.

g I.

h II. 5.

i 14. S.

PRO P. XXVI. THE OR.

F, in a parallelogram, be conftitute another parallelogram fimilar to the whole, and alike fituate, and having an angle common with it, they fhall be about the fame diameter.

Let the parallelogram AF be conftitute in the parallelogram ABCD, fimilar to it, and alike fituate, having the angle ĎAB common to both; then the parallelograms ABCD and AF are about the fame diameter AC.

;

a 24.

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 fame diameter, they will be fimilar to one another 2; and DA to AB as GA to KA; but, becaufe the parallelograms ABCD, GE, are likewife fimilar, DA is to AB as GA is to AE therefore, as GA is to AE fo is GA to AK ; therefore AE is I. s. equal to AK, the greater to the lefs, which is impoffible; 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.

c Hyp.

e 9. 5.

PROP.

PRO P. XXVII. T HEOR.

F all parallelograms applied to the fame right line, and want, ing in figure by parallelograms fimilar and alike fituate to that defcribed on half the line, the great eft is that which is applied to the half line, and fimilar to the defect.

Let AB be a right line bifected in the point C; and let the parallelogram AĎ be applied to the right line AB, wanting in figure by the parallelogram CE, fimilar and alike fituate to that defcribed 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 fimilar and alike fituate to CE. For, let the parallelogram AF be applied to the right line AB, wanting in figure by the parallelogram HK, fimilar and alike fituate to ČE; then the parallelogram AD is greater than AF. For, because the parallelogram CE is fimilar to HK, they will ftand about the fame diameter. Let DB, that diameter, be drawn, and the figure defcribed; then the parallelograms CF, FE, are equal ; add HK, which is common to both; then the whole CH is equal to the whole KE; but CH is equal to GC; 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,

UPON 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 defcribed on half the line, as the defect must be fimilar.

It is required, upon the given right line AB, to apply a paral lelogram equal to the right lined figure C, and deficient by a parallelogram fimilar to D; and the right lined figure C not greater than the parallelogram defcribed on half the line AB, which is fimilar to D. For, bifect AB in E, and on EB defcribe a parallelogram EF fimilar and alike fituate to Da, 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