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logram CF, is the same with the ratio cause as DC to CE, so is the paralle which is compounded of the ratios of logram CH to the parallelogram CF, their sides.
but as DC to CE,
D H Let BC, CG be placed in a straight is L to M; whereline; therefore DC and CE are also fore L is (11. 5.) to in a straight line (14. 1.); and com- M, as the parallelo- B plete the parallelogram DG ; and, gram CH to the pataking any straight line K, make (12. rallelogram
CF : 6.) as BC to CG, so K to L; and as Therefore, since it DC to CE, so make (12. 6.) L to M: has been proved, Therefore, the ratios of K to L, and L that as K to L, so is to M, are the sanie with the ratios of the parallelogram the sides, viz. of BC to CG, and DC AC to the parallelogram CH ; and, as to CE. But the ratio of K to M is L to M, so the parallelogram CH to that which is said to be compounded the parallelogram CF; ex æquali, (22. (A def. 5.) of the ratios of K to L, and 5.) K is to M, as the parallelogram AC L to M: Wherefore also K has to M to the parallelogram CF: but K has the ratio compounded of the ratios of to M the ratio which is compounded the sides : And because as BC to CG, of the ratios of the sides: therefore so is the parailelogram AC to the pa- also the parallelogram AC has to the rallelogram CH (1. 6.) but as BC to parallelogram CF the ratio which is CG, so is K to L; therefore K is (11, compounded of the ratios of the sides. 5.) to L, as the parallelogram AC to Wherefore equiangular parallelothe parallelogram CH: Again, be- grams, &c. Q. E. D.
PROP. XXIV. THEOR.
The parallelograms about the diameter of any parallelogram, are similar
to the whole, and to one another.
Let ABCD be a parallelogram, of site sides of parallelograms are equal which the diameter is AC; and ÉG, to one another, (34. 1.) AB is (7.5.) to HK the parallelograms about the dia- AD, as AE to AG: and DC to CB as meter: The parallelograms EG, HK GF to FE ; and also CD to DA, as FG are similar both to the whole paralle- to GA: Therefore the sides of the palogram ABCD, and to one another. rallelograms ABCD, AEFG about the
Because DC, GF are parallels, the equal angles are angle ANC is equal (29. 1.) to the an- proportionals; and gle AGF: For the same reason, be- they are therefore cause BC, EF are parallels, the angle similar to one anoABC is equal to the angle AEF. And ther(1. def. 6.); For each of the angles BCD, EFG is equal the same reason, to the opposite angle DAB, (34. 1.) the parallelogram ABCD is similar to and therefore are equal to one ano- the parallelogram FHCK. Wherefore ther; wherefore the parallelograms each of the parallelograms GE, KH ABCD, AEFG, are equiangular: And is similar to DB: But the rectilineal because the angle ABC is equal to the figures which are similar to the same angle AEF, and the angle BAC como rectilineal figure, are also similar to mon to the two triangles BAC, EAF, one another (21.6.); therefore the pathey are equiangular to one another; rallelogram GE is similar to KH. therefore (4. 6.) as AB to BC, so is Wherefore the parallelograms, &c. AE to EF: And because the oppo- Q. E. D.
PROP. XXV. PROR.
To describe a rectilineal figure which shall be similar to one, and equal
to another given reclilineal figure. Let ABC be the giver rectilineal upon the second ; therefore as BC to figure, to which the figure to be de- CF, so is the rectilineal figure ABC to scribed is required to be similar, and KGH: But as BC to CF, so is (1. 6.) D that to which it must be equal. It the parallelogram BE to the paralleis required to describe a rectilineal logram EF: Therefore as the rectilifigure similar to ABC, and equal to D. neal figure ABC is to KGH, so is the
Upon the straight line BC describe paralielogram E to the parallelogram (Cor. 45. 1.) the parallelogram BE e EF (11.5.): And the rectilineal figure
. qual to the figure ABC, also upon ABC is equal to the parallelogram BE; CE describe (Cor. 45. 1.) the parallelogram CM equal to D, and having the angle FCE equal to the angle CBL: Therefore BC and CF are in a straight line, (29. 1, 14. 1.) as also LE and EM: Between BC and CF find (13. 6.) a mean proportional GH, and therefore the rectilineal figure KGH upon GH describe (18. 6.) the rectili- is equal (14. 5.) to the parallelogram neal figure KGH similar and similarly EF: But EF is equal to the figure D: situated to the figure ABC: And be- wherefore also KGH is equal to D; cause BC is to GH as GH to CF, and and it is similar to ABC. Therefore if three straight lines be proportionals, the rectilineal figure KGH has been as the first is to the third, so is (2 Cor. described similar to the figure ABC, 30. 6.) the figure upon the first to the and equal to D. Which was to be similar and similarly described figure done.
PROP. XXVI. THEOR.
If two similar parallelograms have a common angle, and be similarly
situated; they are about the same diameter. Let the parallelograms ABCD, are similar to one another (24. 6.); AEFG be similar and siroilarly situ- Wherefore as DA to AB, so is (1. def. ated, and have the angle DAB com- 6.) GA to AK: But because ABCD mon. ABCD and AEFG are about and AEFGare siinilar parallelograms, the same diameter.
as DA is to AB, so is GA to AE; For if not, let,
therefore (11. 5.) as GA to AE, so if possible, the pa
GA to AK; wherefore GA has the rallelogram BD
same ratio to each of the straight lines have its diameter
AE, AK; and conseqnently AK is AHC in a different
equal (9.5.) to AE, the less to the straight line from AF, the diameter greater, which is impossible : Thereof the parallelogram EG, and let GF fore ABD AKHG are not about the meet AHC in H; and through H draw same diameter; wherefore ABCD HK parallel to AD or BC: Therefore and AEFG must be about the same the parallelograms ABCD, AKHG diameter. Therefore, if two similar, being about the same diameter, they &c. Q. E. D.
* To understand the three following in the same angle, propositions more easily, it is to be and between the observed,
same parallels, by 1. That a parallelogram is said the parallelogram to be applied to a straight line, when DC; and DC is it is described upon it as one of its therefore called the defect of AE. sides. Ex. gr. the parallelogram AC 3. And a parallelogram AG is said is said to be applied to the straight to be applied to a straight line AB, line AB.
exceeding by a parallelogram, when * 2. But a parallelogram AE is said AF, the base of AG, is greater than to be applied to a straight line AB, AB; and therefore AG exceeds AC, deficient by a parallelogram, when the parallelogram described upon AB AD, the base of AE, is less than AB; in the same angle, and between the and therefore AE is less than the same parallels, by the parallelogram parallelogram AC described upon AB BG.
PROP. XXVII. THEOR.
Of all parallelograms applied to the same straight line, and deficient by parallelograms, similar and similarly situated to that which is de scribed upon the half of the line; that which is applied to the half, and is similar to its defect, is the grcalest.
Let AB be a straight line divided Because the parallelogram CF is into two equal parts in C and let the equal (43. 1.) to FE, add KH to both, parallelogram AD be applied to the therefore the whole CH is equal to half AC, which is therefore delicient the whole KE: But CH is equal from the parallelogram upon the (86. 1.) to CG, because the base AC whole line AB by the parallelogram is equal to the base_CB; therefore CE upon the other half CB: Of all CG is equal to KE: To each of these the parallelograms applied to any add CF; then the whole AF is equal other parts of AB, and deficient by to the gnomon CHL: Therefore CE, parallelograms that are similar, and or the parallelogram AD, is greater similarly situated to CE, AD is the than the parallelogram AF. greatest.
Next, let AK, the base of AF, be Let AF be any parallelogram ap- less than AC, and the same construcplied to AK, any other part of AB than tion being made the parallelogram the half, so as to be deficient from DH is equal to DG (36. 1.) for HM the parallelogram upon the whole is equal to MG, (34. line AB by the parallelogram KH 1.) because BC is esimilar, and similarly situated to CE; qual to CA; wherefore AD is greater than AF.
DH is greater than First, let AK, the base of AF, be LG: But DH is equal greater than AC the half of AB; and (43. 1.) to DK; there. because CE is similar
fore DK is greater to the parallelogram
than LG: To each of KH, they are about
these add AL; then the whole AD is the same diameter:
greater than the whole AF. There (26. 6.) Draw their
fore, of all parallelograms applied, diameter DB, and
&c. Q. E. D. complete the scheme:
PROP. XXVIII. PROB.
To a given straight line to apply a parallelogram equal to a given recti
lineal figure, and deficient by a parallelogram similar to a given parallelogram: But the given rectilineal figure to which the parallelograin to be applied is to be equal, must not be greater than the parallelogram applied to half of the given line, having its defect similar to the defect of that which is to be applied ; that is, to the given parallelogram.
Let AB be the given straight line, than it; and EF is equal to AG; and C the given rectilineal figure, to therefore EF also is greater than C. which the parallelogram to be applied Make (25. 6.) the parallelogram is required to be equal, which figure KLMN equal to the excess of EF must not be greater than the paral- above C, and similar and similarly lelogram applied to the half of the situated to D; but D is similar to line having its defe ut from that upon EF, therefore (26. 1.) also KM is the whole line similar to the defect of similar to EF: Let KL be the homothat which is to be applied ; and let logous side to EG, and LM to GF: D be the parallelogram to which this And because EF is equal to C and defect is required to be similar. It is KM together, EF is greater than required to apply a parallelogram to KM; therefore the straight line EG G OF
is greater than KL, and GF than LM: Make GX equal to LK, and GO equal to LM, and complete the parallelogram XGOP; Therefore XO is equal and similar to KM; but KM is similar to EF: wherefore also XO is similar to EF, and therefore XO and EF are about the same dia. ineter : (26. 6.) Let GPB be their
diameter, and complete the scheme : the straight line AB, which shall be Then because EF is equal to C and equal to the figure C, and be deficient KM together, and XOʻa part of the from the parallelogram upon the one is equal to KM a part of the whole line by a parallelogram similar other, the remainder, viz. the gnomon to D.
ERO, is equal to the remainder C: Divide AB into two equal parts And because OR is equal (34. 1.) to (10. 1.) in the point E, and upon EB XS, by adding SR to each, the whole describe the parallelogram EBFG OB is equal to the whole XB: But similar (18. 6.) and similarly situated XB is equal (36.) to TE, because the to D, and complete the parallelogram base AE is equal to the base EB; AG, which must either be equal to wherefore also TE is equal to OB: C, or greater than it, by the deter- Add XS to each, then the whole TS mination : And if AG be equal to C, is equal to the whole, viz. to the then what was required is already gnomon ERO: But it has been provo done : For, upon the straight line AB, ed that the gnomon ERO is equal to the parallelogram AG is applied equal C, and therefore also TS is equal to to the figure C, and deficient by the C. Wherefore the parallelogram TS, parallelogram ÉF similar to D : But, equal to the given rectilineal figure if AG be not equal to C, it is greater C, is applied to the given straight
line AB deficient by the parallelogram cause SR is similar to EF. (24. 6.) SR, similar to the given one D, be- Which was to be done.
PROP. XXIX. PROBLEM.
To a given straight line to apply a parallelogram equal to a given rectie
lineal figure, exceeding by a parallelogram similar to another given. Let AB be the given straight line, (26. 6.) Draw their diameter FX, and C the given rectilineal figure to and complete the scheme. Therefore, which the parallelogram to be applied is required to be equal, and D the parallelogram to which the excess of the one to be applied above that upon the given line is required to be similar. It is required to apply a parallelogram
LM to the given straight line AB which shall be equal to the figure C, exceeding by a parallelogram similar to D.
PX Divide AB into two equal parts in since GH is equal to EL and C tothe point E, and upon EB describe gether, and that GH is equal to MN; (18.) the parallelogram EL similar, MN is equal to EL and C: Take and similarly situated to D: And away the common part EL;
then make (25. 6.) the parallelogram GH the remainder, viz. the gnomon NOL, equal to EL and C together, and is equal to C. And because AE is similar, and siunilarly situated to D; equal to EB, the parallelogram AN wherefore GH is similar to EL: (21. is equal (36. 1.) to the parallelogram 6.) Let KH be the side homologous NB, that is to BM. (43. 1.) Add to FL, and KG to FE: And because NO to each ; therefore the whole, viz. the parallelogram GH is greater than the parallelogram AX, is equal to EL, therefore the side KH is greater the gnomon NOL. But the gnomon than FL, and KG than FE: Produce NOL is equal to C; therefore also FL and FE, and make FLM equal to AX is equal to C. Wherefore to the KH, and FEN to KG, and complete the straight line AB there is applied the parallelogram MN. MN is therefore parallelogram AX equal to the given equal and similar to GH; but GH is rectilineal C, exceeding by the pare similar to EL; wherefore MN is si- allelogram PO, which is similar to D, milar to EL, and consequently EL because PO is similar to EL (24. 6.) and MN are about the same diameter: Which was to be done.
PROP. XXX. PROBLEM.
To cut a giren straight line in extreme and mean ratio. Let AB be the given straight line: ing by the figure AD similar to BC: it is required to cut it in extreme aud (29. 6.) But BC is a square, therefore mean ratio.
also AD is a square; and because Upon AB describe (46. 1.) the BC is equal to CD, by taking the square BC, and to AC apply the pare common part CE from each, the reallelogram CD equal to BĆ, exceed- mainder BF is equal to the remainder