quently [by g. 5.) the right lined figures NH, SR are equal : They are also similar and alike situate: Therefore [by the following lemma] G H is equal to Pr. And because AB is to CD, as E F is to PR, and PR is equal to G H (by 7. 5.] A B will be to CD, as Ek is to GH. If therefore four right lines be proportional, the right lined figures that are described upon them fimilar and alike situate, will be also proportional: and it right lined figures described upon four right lines similar and alike fituate, be : proportional; these right lines will be proportional Which was to be demonstrated. #1 L E M M A. But we shall thus demonstrate, that the homologous fides of fimilar and equal right lined figures, are equal. Let the right lined figures NH, SR be fimilar and equal : and let HG be to GN, as r p is to ps: I say. RP is equal to HG. - For if they be unequal, one of them will be the greater. Let this be (RP. Then because Rp is to Ps, as hy is to GN: and alternately [by 16. 5.] as Rp is to GH, fo is ps to GN. But PR is greater than GH: therefore PS will be greater than GN: Wherefore [by 20. 6.) the right lined figure r s is greater than the right lined figure HN; and also equal; which cannot be : Therefore Pr is not unequal to gh: Wherefore it is equal to it. Which was to be demonstrated. i This proposition is always true; if all the four figures be similar and alike fituate. But when only two and two are so : it will not always be true, unless the two similar and alike fituate figures be both described upon those two righe lines that are the antecedents and consequents of the equal ratios ; as if the right line a be to B, as c is to D, and the figures C, H described upon the first and fourth right lines A and o be similar and alike fi. tuate, and both the figures 1, H described G K upon the second and А B o D) third right lines' B, c be both similar and alike fituate, there four figures G, I, K, I will not be proportional, as i; easily apprehended, from the bare contemplation of the figurcs. Should K Should not the lemma to this proposition have been made a proposition, and set down before this proposition? PROP. XXIII. THEOR. Equiangular parallelograms are to one another in a ratio compounded of the ratios of their fides k. Let A C, C F be equiangular parallelograms, having the angle BCD equal to the angle eco: I say the parallelogram A c is to the parallelogram' ct in the ratio compounded of the ratios of the sides : that is, of the ratio of BC to CG, and of the ratio of DC to ce. For put BC, cg in the same right line. Then [by 14. 1.] dc will be in the same right line with c E i and compleat the parallelogram DG, also let k be any right line, and [by 22. 6.) make as BC to CG, fo is K to , and as DC is to CE, fo is 1 to M. Then the ratios of K to L, and I to M are the fame as the ratios of the fides, viz. that of BC to cG, and of DC to ce: But [by 5. def. 6.) the ratio of K to M is compounded of the ratio of K to L, and of the ratio of 1, to M. Wherefore the ratio of K tom is I! the ratio compounded of the ratios of the sides. And because [by 1. 6.] as Bc is to cG, so is the paralleloB gram A G to the parallelogram CH: But as BC to cG, so is k to 1r Therefore [by 11. 5.] as k is to L, so is the parallelogram A c to the parallelogram ch. Again, because E F [by 1.6.] as Dc is to CE, fo is the parallelogram ch, to the paralleloKIM gram cf; and as Dc is to C E, so is I to M: Therefore as L is to M, so [by 11. 5.) will the parallelogram ch be to the parallelgram cf. And so fince it has been proved that as k is. to L, so is the parallelogram Ac to the parallelogram ch: But as I to M, so is the parallelogram ch to the parallelogram cf: it will be, by equality [by 22. 5.) as K is to M, so is the parallelogram Ac to the parallelogram C F. But the ratio of K to m is that compounded of the ratios of the sides: Therefore the ratio of the parallelogram AC to A to the parallelogram cf is compounded of the ratios of their fides. Therefore equiangular parallelograms are to one another in a ratio compounded of the ratios of their fides. Which was to be demonstrated. B k Some may perhaps like the following demonftration of this proposition better than Euclid's. It will be (by 1. 6.) as B C to cã, fo is the parallelogram AC to the parallelogram ch. And as Bc is to CG, so is the rectangle under BC and cd to the rectangle under cg and cd. Wherefore [by equality] as the rectangle under Bc and cd is to the rectangle under c G and CD, so is the parallelogram ac to the parallelogram ch. Therefore (by alternation] as the "rectangle under BC and CD, is to the parallelogram A C, so is the rectangle ander ca and CD, to the parallelogram cn. For the same reason, the rectangle ander ce and cc is to the parallelogram CF, as the reciangle under CD and co is to the parallelogram ch. Where fore [by equality) as the rectangle under Bc and CD, is to the parallelogram A C, so is the rectangle under ce and co to the parallelogram cf. Therefore (by alternation) as the rectangle under B C and c D is to the rectangle under ce and cG, so is the parallelogram Ac, to the parallelogram CF. W. W. D. If there be two triangles having one angle of the one equal to one angle of the ocher : those triangles will be to one another in the ratios compounded of the ratio of the sides containing the equal angles. This will appear evident by supposing the angles B, D, E, of the equiangular parallelograms AB CD, E CGF to be joined by right lines. For then because [by 34. 1.) the triangles BDC, ECG are each the one half of the said parallelograms ; fince halves are as the wholes, and since the vertical angles of these triangles at c are equal to one another. Therefore [by this prop. ] the two criangles at BD C, ECD will be to one another in a ratio compounded of the ratios of their fides : and fo [by def. 5. 6.] one triangle will be to another, as the rectangle under the sides containing an angle of the one, is to the re&angle under the fides containing an equal angle of the other. This theorem is also easily demonstrated. If the diagonals of two trapeziums interfect one another in a given angle, those trapeziums will be to one another in a ratio compounded of the ratios of the diagonals. PRO P. XXIV. THEOR. The two parallelograms that stand about the diameter of every parallelogram, are similar to one another, and to the whole parallelogram! Let ABCD be a parallelogram, whose diameter is A c, and let EG, HK be parallelograms standing about the diaA E B meter Ac: I fay the parallelo grams EG, HK are similar to one G F another, and to the whole paralH lelogram ABCD. For because E F is drawn parallel to one side bc of the trian. gle ABC; [by 2. 6.] it will be D K C as BE to En, so is cF to FA. Again, because FG is drawn parallel to the side cd of the triangle A CD, as cF is to FA, fo will do be to ga. But it has been proved that as CF is to FA, fo is be to EA: Therefore [by 11. 5.) as be is to E A, fo is DG to IGA: and by compounding. [by 18. 5.] as B A is to Aė, fo is DA to AG; and inversely [by 16. 5.) as BA is to AD, so is A É to AG: Therefore the sides of the parallelograms ABCD, EG about the common angle B A D are proportional. And because GF is parallel to DC, the angle AGF [by 29. 1.) is equal to the angle ADC. But the angle G FA is equal to the angle DCA, and the angle DAC is common to each of the triangles A DC, AGF: Therefore the triangle ADC will be equiangular to the triangle AGF. By the fame reason the triangle ABC will be equiangular to the triangle Afe: Therefore the whole parallelogram A BCD is equiangular to the parallelogram EG: Wherefore [by 4. 6.] as A D is to D'c, so is AG to GF. But as Dc is to ca, fo is G F to Fa, and as Ac is so is AF to fe; and also as CB is to BA, fo is FE to E A: Wherefore because it has been proved that DC is to c A, as G F is to FA, and as a c is to CB, so is AF to FE; it will be, by equality, [by 22. 5.] as Dc is to CB, so is G F to FE: Therefore the fides of the parallelograms A B C D, E G which are about the equal angles, are proportional; and accordingly [by 1. def. 6.] the parallelogram ABCD is similar to the parallelogram E G. By the to CB, the same reason the parallelogram abcd is also similar to the parallelogram KH: Therefore the parallelograms EG, H K are each of them fimilar to the parallelogram ABCD. But [by 21. 6.) those right lined figures which are similar to the same right lined figure, are similar to one another : Therefore the parallelogram EG is similar to the parallelogram HK. Wherefore the parallelograms that stand about the dia-. meter of every parallelogram are similar to one another, and to the whole parallelogram. Which was to be demonstrated. ! Some have observed here that the parallelograms about the diameter are fimilar figures, having their fides directly proportional, and the complements are equal parallelograms, having their fides reciprocally proportional to one another : Also either of the compliments is a mean proportional between the parallelograms about the diameter. And these are to one another in che duplicate ratio of their homologous fides. PROP. XXV. PROBL. To describe a right lined figure fimilar to one given right lined figure, and equal to anothers Let A B C be the given right lined figure to which the figure is to be made fimilar, and the right lined figure to which it is to be made equal : it is required to make a right lined figure similar to ABC, and equal to D. For [by 44. and 45. 1.] to the right line BC apply the parallelogram be equal to the triangle ABC, and to the right line ce apply the parallelogram cm equal to D, and having the angle FCE, equal to the angle CBL: Then [by 14. 1.] BC, CF are both in one right line, as also LE, EM. А D Now [by 13. 6.] к find nd a mean proportional gh between С BC, CF, and [by B , 18. 6.] upon GH H describe the right L E M lined figure KGH T 2 fimilar |