one another : (16. 6.). Therefore the triangle ABC to the triangle ABG. triangle ABG is equal to the triangle Therefore the triangle ABC has to DEF: And because as BC is to EF, the triangle ABG the duplicate ratio 80 EF to BG; and that if three of that which BC hac tó EF: But straight lines be proportionals, the the triangle ABG is equal to the tri angle DEF; wherefore also the triangle ABC has to the triangle DEF the duplicate ratio of that which BC has to EF. Therefore similar tri. angles, &c. Q. E. D. BG Cor. From this it is manifest, that first is said (10 def. 6.). to have to if three straight lines be proportionals, the third the duplicate ratio of that as the first is to the third, so is any which it has to the second ; BC triangle upon the first to a similar, therefore has to BG the duplicate and similarly described triangle upon ratio of that which BC has to EF: the cecond. But as BC to BG, so is (1. 6.) the À E PROP, XX. THEOR. Similar polygons may be divided into the same number of similar tri angles, having the same ratio to one another that the polygons huve ; and the polygons have to one another the duplicate ratio of that which their homologous sides have. Let ABCDE, FGHKL be similar 6.) to the whole angle FGH; therepolygons, and let AB be the homo- fore the remaining angle ERC is logous side to FG: The polygons equal to the remaining angle LGH: ABCDE, FGHKL may be divided And because the triangles ABE, into the same number of similar tri- FGL are similar, EB is to BA, as angles, whereof each to each has the LG to GF; (1. def. 6.) and also, besame ratio which the polygons have; cause the polygons are similar, AB is and the polygon ABCDE has to the to BC, as FG to GH, (1. def. 6.) polygon FGHKL the duplicate ratio therefore, ex æquali, (22. 5.) EB is to of that which the side AB has to the BC, as LG to GH ; that is, the sides side FG. about the equal angles EBC, LGH Join BE, EC, GL, LH: And be- are proportionals; therefore, (22. 5.) cause the polygon ABCDE is similar the triangle EBC is equiangular to to the polygon FGHKL, the angle the triangle LGH, and similar to it. BAE is equal to the angle GFL, (1, (4. 6.) For the same reason, the tridef. 6.) and BA is to AE, as GF to FL: (1. def. 6.) Wherefore, because the triangles ÁBE, FGL, have an angle in one equal to an angle in the other, and their sides about these equal angles proportionals, the triangle A BE is equiangular, (6. 6.) and therefore similar to the triangle PGL: angle ECD likewise is similar to the (4. 6.). Wherefore the angle ABE triangle LAK: therefore, the similar is equal to the angle FGL: And polygons ABCDE, FGHKL are di ecause the polygons are similar, vided into the samne number of similar the whole angle ABC is equal (1. def. triangles. A M E BL K Also, these triangles have, each to 5.), Wherefore, as the triangle ABE each, the same ratio which the poly- to the triangle FGL, so is the polygon gons have to one another, the ante. ABCDE to the polygon FGHKL : cedents being ABE, EBC, ECD, and But the triangle ÅBE has to the trithe consequents FGL, LGH, LHK: angle FGL, the duplicate ratio of that And the polygon ABCDE has to the which the side AB has to the homolo polygon FGHKL the duplicate ratio gous side FG. Therefore, also the of that which the side AB has to the polygon ABCDE has to the polygon homologous side FG. FGHKI the duplicate ratio of that Because the triangle ABE is similar which AB has to ihe bomologous side to the triangle FGL, ABE has 10 FGL FG. Wherefore siznilar polygons, the duplicate ratio (19. 6.) of that &c. Q. E. D. which the side BE has to the side GL: Cor. 1. In like manger, it may be For the same reason, the triangle BEC proved, that similar four sided figures, has to GLH the duplicate ratio of that or of any number of sides, are one to which BE has to GL: Therefore, as another in the duplicate ratio of their the triangle ABE to the triangle FGL, homologoas sides, and it has already so (11. 5.) is the triangle BEC to the been proved in triangles. Therefore, triangle GLH. Again, because the universally, similar rectilineal figures triangle EBC is similar to the triangle are to one another in the duplicate raLGH, EBC has to LGH the duplicate tio of their homologous sides. ratio of that which the side EC has to Cor. 2. And if to AB, FG, two of the side LH: For the same reason, the homologous sides, a third proporthe triangle ECD has to the triangle tional M be taken, AB has (10. def. 5.) LHK, the duplicate ratio of that to M the duplicate ratio o: that which which EC has to LH: As therefore AB has to FG: But the four sided the triangle EBC to the triangle LGH, figure or polygon upon AB, has to the so is (11. 5.) the triangle ECD to the four sided ligure, or polygo: upon triangle LHK: But it has been FG, likewise the duplicate ratio of proved, that the triangle EBC is like- that which AB has to FG: Therefore, wise to the triangle LGH, as the tri as AB is to M, so is the figure upon angle ABE to the triangle FGL. AB to the figure upon FG, which was Therefore, as the triangle ABE is to also proved in triangles. (Cor. 19.6.) the triangle FGL, so is the triangle Therefore, universally, it is manifest, EBC to triangle LGH, and triangle that if three straight lines be proporECD to triangle LHK: And there- tionals, as the first is to the third, so fore, as one of the antecedents to one is any rectilineal figure upon the first, of the consequents, so are all the an to a similar and similarly described tecedents to all the consequents. (12. rectilineal figure upon the second. PROP. XXI, THEOR. Rectilineal figures, which are similar to the same rectilineal figure, are also similar to one another. Let each of the rectilineal figures about the equal A, B be similar to the rectilineal figure angles proporC: The figure A is similar to the tionals. (1. Def. figure B. 6.) Again, beBecause A is similar to C, they are cause B is siequiangular, and also have their sides milar to C, they are equiangular, and have their sides C proportionals. Wherefore the recabout the equal angles proportion- tilineal figures A and B are equials (1. Def. 6.): Therefore the figures angular, (1. Ax. 1.) and have their A, B, are each of them equiangular sides about the equal angles proporto C, and have the sides about the tionals. (11. 5.) Therefore A is simiequal angles of each of them and of lar (1. Def. 6.) to B. Q. E. D. PROP. XXII. THEOR. K B M N G H Р If four straight lines be proportionals, the similar rectilineal figures similarly described upon them shall also be proportionals; and if the similar rectilineal figures similarly described upon four straight lines be proportionals, those straight lines shall be proportionals. Let the four straight lines AB, CD, situated to either of the figures MF, EF, GH be proportionals, viz. AB to NH: Then, because as AB to CD, só CD, as EF to GH, and upon AB, CD let the similar rectilineal figures KAB, LCD be similarly described ; and upon EF, GH the similar rectilineal figures х MF, NH in like manner: The rectilineal figure KAB is to LCD, as MF to NH. To AB, CD take a third proportional (11. 6.) X; and to EF, GH, a R third proportional 0: And because is EF to PR, and that upon AB, CD AB is to CD, as EF to GH, and that are described the similar and similarly CD is (11. 5.) to X, as GH to 0; situated rectilineals KAB, LCD, and wherefore, ex æquali, (22. 5.) as AB upon EF, PR, in like manner, the sito X, so EF to O: But as AB to X, milar rectilineals MF, SR; KAB is to so is (2 Cor. 20. 6.) the rectilineal KAB LCD, as MF to SR; but, by the hyto the rectilineal LCD; and as EF to pothesis, KAB is to LCD, as MF to 0, so is (2 Cor. 2. 6.) the rectilineal NH; and therefore the rectilineal MF MF to the rectilineal NH: There- having the same ratio to each of the fore, as KAB to LCD so (11. 5.) is two ÑH, SR, these are equal (9. 5.) MF to NH. to one another: They are also similar, And if the rectilineal KAB be to and similarly situated; therefore GII LCD, as MF to NH; the straight is equal to PR: And because as AB line AB is to CD, as EF to GH. to CD, so is EF to PR, and that PR Make (12. 6.) as AB to CD, so EF is equal to GH; AB is to CD, as EF to PR, and upon PR describe the rec to GH. If therefore four straight tilineal figure SR similar and similarly lines, &c. Q. E. D. PROP, XXIII. THEOR. Equiangular parallelograms have to one another the ratio which is comi pounded of the ratios of their sides. Let AC, CF be equiangular parale equal to the angle ECG: the ratio of lelograms, having the angle BCD the parallelogram AC to the paralle E F KLM logram CF, is the same with the ratio cause as DC to CE, so is the parallewhich is compounded of the ratios of logram CH to the parallelogram CF. their sides. but as DC to CE, so он 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 parallelogramDG; 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 paralelogram 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. н Let ABCD be a parallelogram, of site sides of parallelograms are equal which the diameter is AC; and EG, 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 com- 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. D K PROP. XXV. PROB. 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. M G H A D 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 similarly situ- Wherefore as DA to AB, so is (1. def. ated, and have the angle DAB com- 6.) GA IN AK: But because ABCD mon. ABCD and AEFG are about and AEFG are 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. K |