PROP. XVII. THEOR. if three right lines be proportional : the rectangle contained under the extremes is equal to the square of the mean line : and if the rectangle contained under the extremes be equal to the square of the mean line, the three right lines will be proportional. B Let the three right lines A, B, C be proportional ; so that a be to B, as B is to c: I say the rectangle contained under the right lines A, c is equal to the square of the mean line B. For let D be put equal to B. Then because as a is to B, so is B to c, and B is equal to D; it will be as A is to B, so is D to c. But if four right lines be proportional, А the rectangle contained under the extremes [by 16. 6.] is equal to the rectangle contained under the means : Therefore the rectangle с contained under the right A B lines A, C, is equal to the rectangle contained under the right lines B, D. rectangle contained under the right lines B, D is equal to the square of B; for B is equal to D: Therefore the rectangle contained under the right lines A, c is equal to the square of B. But now let the rectangle contained under the right lines A, c be equal to the square of B: I say it will be as A is to B, so is B to c. For the same construction remaining, because the rectangle contained under the right lines A, C, is equal to the square of B; but the square of B is the rectangle contained under B, D ; for B is equal to D, the rectangle contained under the right lines A, c is equal to the rectangle contained under the right lines B, D, But if the rectangle contained under the extremes be equal to the rectangle contained under the means ; [by 16. 6.) the four right lines are proportional : Therefore as A is to B, fo is D S3 to c: But B is equal to D: Wherefore as A is to B, so is B to c. If therefore three right lines be proportional, the rectangle contained under the extremes will be equal to the square of the mean line: and if the rectangle contained under the extremes be equal to the square of the mean line, the three right lines will be proportional. Which was to be demonstrated. PROP. XVIII. THEOR. Upon a given right line, to describe a right-lined fi gure similar and alike ftuate to a given right-lined figure Let A B be the given right line, and ce the given rightlined figure : it is required to describe a right-line figure upon A B fimilar and alike situate to the given right lined figure ce. Join Df; and (by 23, 1.) at the right line A B, and at the points A, B in it, make the angle G A B equal to the angle c, and the-angle A B G equal to the angle cdf: Then [by 32. 1.] the angle remaining CFD is equal to the angle remaining AGB: Therefore, the triangle FCD is equiangular to the triangle GAB; and accordingly [by 4. 6.) as FD is to GB, so is Fc to G A and CD to A B, Again, at the right line BG, and at the points B, G in it, make the angle BG H equal to the angle dfe, and the angle G BH equal to the angle FDE: Then the remaining angle at E is equal to the remaining angle at H: Therefore the triangle FDE is equiangular to the triangle GBH: and so [by 4. 6.] aś Fd is to GB, fo is F E to GH and ED TO HB. But it has been also proved, that as FD is to GB, fo is FC to GA and CD to AB: and therefore [by 11. 5.) as Fc is to GA, fo is c D to A B and F'E to GH, and moreover ED to H B. Wherefore because the angle CFD [by construction) is equal to the angle A GB, and the angle DFE to the angle BGH: the whole angle cfe is equal to the whole angle A GH. By the same realon the angle CDE is also equal to the angle A BH, and besides the angle at c is equal to the angle at A; but the angle at e is equal to the angle at H; therefore Ah is equiangular to ce, and it has the sides about the equal angles proportional : Wherefore [by 1. def. 6.] the right lined figure A H will be similar to the right lined figure CE. Therefore upon the given right line A B the right lined figure Ah is described fimilar to the given right lined figure c E, and alike situate. Which was to be done. - Similar right lined figures are said to be alike described, or set upon right lines, when those right lines are the homologous fides of the similar right lined figures; or when the equal angles are constituted upon those right lines, and the remaining equal angles, and proportional sides of the figures always follow one another in order As the triangles ABC, D E F are not only fimilar but are alike fituate upon the right lines BC, EF, when the an. D gles B, C are equal to the angles A E, P; and it be as A B to BC, so is De to EF, &c. But they would not have been fimilarly fituate, ifB CE the angle B had been equal to F, and c equal to B. In like man A B I K ner the rectang'es A C, E G are fimilar and alike fituate upon D c the right lines DC, HG, when AD is to do as E H is to HG E HG &c. But the rectangles A C, 1G are said not to be similar situate upon the right lines DC, HG although they are fimilar, as is manifelt. But notwithstanding the same figures will be alike Situate or described upon the righị lines DC, 1H, or upon AD, H, That the figures be alike situate, is a necessary condition. For if they be not, this propofition, and all the o hers following, limited to that condition will not be always true, because two Similar figures may be described upon the same right line not aliko H SA B E alike fituate, that will be uno В çqual, as the triangles ABC, DE F, (where ac is equal to DF, and the angles A, D ; R, А C D F are respectively equal to one another) will be unequal. Note also that all equilateral and equiangular figures are alike described upon any right lines. a PRO P. XIX. THEOR. Similar triangles are to one another in the duplicate ratio of their bomologous fides. Let the trianges ABC, D E F be similar, baving the angle B equal to the angle 'e ; and let A B be to B Ç, as DE is to Ę F: so that [by. 12. def. 5.). the side BC is homologous to the side EF: I say the triangle A B C is to the triangle Def in the duplicate ratio of that of the fide BC to the fide e F. For [by 11. 6.] find a third proportional BG to BC, EF, fo that BC be to EF, as EF is to BG ; and join GA. А Then because it is as A B to B C, so is D E to EF; it will be alternately D [by 16. 5.) as A B is to DE, fo is Bc to E F. But as BC is to E F, so is EF to BG; and therefore [bý II. 5.) as AB is to DE, fo is E F to B'G: Therefore BG СЕ the fides about the equal angles of the triangles A B G, DE F, are reciprocally proportional. But those triangles having one angle of the one equal to one angle of the other, and the sides about the equal angles reciprocally proportional, are [by 15.6.] equal to one another : there fore the triangle ABG is equal to the triangle D E F. And because Bc is to E F, as E'F is to BG; and if three right lines be proportional, the ratio of the first to the third is (by 10. def. s.] duplicate to the ratio that the first has to the second : the ratio of BC to BG will be duplicate of the ratio of BC to E F. But as B C is to B G, so (by 1. 6.] is the triangle ABC to the triangle ABG: Therefore also the the ratio of the triangle A B C to the triangle ABG is duplicate of the ratio of BC to E F. But the triangle ABG is equal to the triangle Dep: and therefore [by 7. 5.] the ratio of the triangle A BC to the triangle D E F is the duplicate of the ratio of BC to EF. Therefore fimilar triangles are to one another in the duplicate ratio of their homologous fides. Which was to be demonstrated. Corollary. From hence it is manifeft, if three right lines be proportional, as the first is to the third, fo is a triangle described upon the first, to a triangle similar and alike situated, described upon the second : because it has been proved as CB is to BG, so is the triangle A BC to the triangle A BG, that is, to the triangle D E F. PROP. XX. THEOR. Similar polygons are divided into equal numbers fimilar triangles bomologous to the wholes; and one polygon bas to another polygon a duplicate ratio to that which one homologous fide of the one bas to an homologous side of the other. Let the polygons A B C DE, FGH KL be fimilar, and \et A B, FG be homologous fides : I say the polygons ABCDE, FGHKL are divided into an equal number of similar triangles, homologous to the wholes; and the polygon ABC D E to the polygon FGHKL has a duplicate fatio to that which one homologous side A B has to another F G Join BE, EC, GL, LH. Then because the A polygon A B C D E is fimilar to the polygon F G HKL; F the angle BAE is G equal to the angle N G FL: But it isE as B A to AE, fo is G F to FL: Therefore because A B E, F G ļ are two triangles hava ск H ing B M D |