Е с DB "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 situaled 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 greatest. 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 AKC KH, they are about these add AL; then the whole AD is the same diameter: F11 greater than the whole AF. Therea (26. 6.) Draw their fore, of all parallelograms applied, diameter DB, and A &c. Q. E. D. CKB complete the scheme: GFM H L D DLE G PROP. XXVIII. PROB. H 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 parallclograin 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 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 prove 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 T R A E SB L M M к N to D. 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. given recti D F B N To a given straight line to apply a parallelogram equal to a 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 to the given straight line AB which shall be equal to the figure C, exceeding by a parallelogram similar to PX D. 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 ; 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 rectilineal C, exceeding by the parsimilar to EL; wherefore MN is sie 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. then PROP. XXX. PROBLEM. To cut a given 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 par- common part CE from each, the reallelogram CD equal to BC, exceed- mainder BF is equal to the remainder AD: And these figures are equian Otherwise, gular; therefore their sides about the Let AB be the given straight line ; equal angles are reciprocally propor. it is required to cut it in extreme and tional: (14. 6.) Wherefore, as FE to mean ratio. ED, so À E to EB: But FE is equal Divide AB in the point C, so that to AC, (34. 1.) that is to AB; and the rectangle contained by AB, BC ED is equal to AE: be equal to the square of AC:(11. 2.) Therefore as BA to AE, Then, because the rect. so is AE to AB: Bit angle AB, BC is equal с в AB is greater than AE; to the square of AC, as wherefore AE is greater BA to AC so is AC to CB: (17. 6.) than EB:(14.5.) There Therefore AB is cut in the extreme fore the straight line AB and mean ratio in C. (3. def. 6.) is cut in extreme and Which was to be done. mean ratio in E. (3. def. 6.) Which was to be done. PROP. XXXI. THEOREM. B upon BA: B D с In right angled triangles, the rectilineal figure described upon the side opposite to the right angle, is equal to the similar and similarly described figures upon the sides containing the right angle. Let ABC be a right angled triangle, CB to BD, so is having the right angle BAC: The the figure upon rectilineal figure described upon BC CB to the similar is equal to the similar, and similarly and similarly dedescribed figures upon BA, AC. scribed figure Draw the perpendicular AD; there And, fore, because in the right angled tria inversely, (B. 5.) angle ABC, AD is drawn from the as DB to BC, so right angle at A perpendicular to the is the figure upon BA to that upon base BC, the triangles ABD, ADC BC: For the same reason, as DC to are similar to the whole triangle ABC, CB, so is the figure upon CA to that and to one another, (8. 6.) and be. upon CB. Wherefore as BD and cause the triangle ABC is similar to DC toge:her to BC, so are the figures ABD, as CB to BA, so is BA to BD; upon BA, AC to that upon BC: (4. 6.) and because these three (24. 5.) But BD and DC together are straight lines are proportionals, as equal to BC. Therefore the figure the first to the third, so is the figure described on BC is equal (A. 5.) to upon the first to the similar, and the similar and similarly described similarly described figure upon the figures on BA, AC. Wherefore, in second: (2 Cor. 20. 6.) Therefore as right angled triangles, &c. Q. E. D. PROP. XXXII. THEOREM. If two triangles which have two sides of the one proportional to iwo sides of the other be joined at one angle, so as to have their homologous sides parallel to one another ; the remaining sides shall be in a straight line. Let ABC, DCE be two triangles proportional to the two CD, DE, viz. which have the two sides BA, AC BA to AC, as CD to DE; ard let AB D be parallel to DC, and AC to DE, the angle BAC was proved to be equal BC and CE are in a straight line. to ACD: Therefore the whole angle Because AB is parallel to DC, and ACE is equal to the two angles ABC, the straight line ACA BAC; add the common angle ACB, meets them, the al then the angles ACE, ACB are equal ternate angles BAC to the angles ABC, BAC, ACB: But ACD are equal ; (29 ABC, BA ACB are equal to two 1.) for the same rea right angles ; (32. 1.) therefore also son, the angle CDEB the angles ACE, ACB are equal to is equal to the angle ACD; where two right angles : And since at the fore also BAC is equal to CDE: And point C, in the straight line AC, the because the triangles ABC, DCE have two straight lines BC, CE, which are one angle at A equal to one at D, on the opposite sides of it, make the and the sides about these angles pro- adjacent angles ACE, ACB equal to portionals, viz. BA to AC, as CD to two right angles; therefore (14. 1.) DE, the triangle ABC is equiangular BC and CE are in a straight line. (6. 6.) to DCE: Therefore the angle Wherefore, if two triangles, &c. Q. ABC is equal to the angle DCE: And E. D. PROP. XXXIII. THEOREM. In equal circles, angles, whether at the centres or circumferences, have the same ratio which the circumferences on which they stand have to one another: So also have the sectors. Let ABC, DEF be equal circles; EHN: and if the circumference BL and at their centres the angles BGC, be greater than EN, likewise the EHF, and the angles BAC, EDF at angle BGL is greater than EHN; their circumferences ; as their circum- and if less, less: there being then ference BC to the circumference EF, four magnitudes, the two circumferso is the angle BGC to the angle EHF, ence BC, EF, and the two angles and the angle BAC to the angle EDF; BGC, EHF; of the circumference BC, and also the sector BGC to the sector and of the angle BGC, have been EHF. taken any equimultiples whatever, Take any number of circumferences viz. the circumference BL, and the CK, KL, each equal to BC, and any angle BGL; and of the circumfernumber whatever FM, MN, each ence EF, and of the angle EHF, any equal to EF: And join GK, GL, HM, HN. Because the circumferences BC, CK, KL, are all equal, the angles A BGC, CGK, KGL are also all equal : (27. 3.) Therefore what multiple soever the circumference BL is of the circumference BC, the same multiple is the angle BGL of the angle BGC: в с E F For the same reason, whatever mul. tiple the circumference EN is of the equimultiples whatever, viz. the citcircumference EF, the same multiple cumference EN, and the angle EHN: is the angle EHN of the angle EHF: And it has been proved, that, if the And if the circumference BL be equal circumference BL begreater than EN, to the circumference EN, the angle the angle BGL is greater than EHN; BGL is also equal (27. 3.) to the angle and if equal, equal; and if less, less: D H N K к M |