equal to the remaining angle GHB, and the triangle FDE equiangular to the triangle GBH: then, because the angle AGB is equal to the angle CFD, BGH to DFE the whole angle AGH is equal to the whole CFE: for the same reason, the angle ABH is equal to the angle CDE; also the angle at A is equal to the angle at C, and the angle GHB to FED; There fore the rectilineal figure ABHC is equiangular to CDEF: but likewise these figures have their sides about the equal angles proportionals: for the triangles GAB, FCD being equiangular, BA: AG: DC: CF (4. 6.); for the same reason, AG: GB CF: FD; and because of the equiangular triangles BGH, DFE, GB: GH:: FD FE; therefore, ex æquali (22. 5.) AG: GH:: CF: FE. In the same manner, it may be proved, that AB: BH:: CD: DE. Also (4. 6.), GH: HB FE: ED. Wherefore, because the rectili neal figures ABHG, CDEF are equiangular, and have their sides about the equal angles proportionals, they are similar to one another (def. I. 6.). Next, Let it be required to describe upon a given straight line AB, a rectilineal figure similar, and similarly situated to the rectilineal figure CDKEF. Join DE, and upon the given straight line AB describe the rectilineal figure ABHG similar, and similarly situated to the quadrilateral figure CDEF, by the former case; and at the points B, H in the straight line BH, make the angle HBL equal to the angle EDK, and the angle BHL equal to the angle DEK; therefore the remaining angle at K is equal to the remaining angle at L; and because the figures ABHG, CDEF are similar, the angle GHB is equal to the angle FED, and BHL is equal to DEK; wherefore the whole angle GHL is equal to the whole angle FEK; for the same reason the angle ABL is equal to the angle CDK: therefore the five-sided figures AGHLB, CFEKD are equiangular; and because the figures AGHB, CFED are similar, GH is to HB as FE to ED; and as HB to HL, so is ED to EK (4. 6.); therefore, ex æquali (22. 5.), GH is to HL, as FE to EK: for the same reason, AB is to BL, as CD to DK: and BL is to LH, as (4. 6.) DK to KE, because the triangles BLH, DKE are equiangular therefore, because the five-sided figures AGHLB. CFEKD are equiangular, and have their sides about the equal angles pro portionals, they are similar to one another; and in the same manner a rec tilineal figure of six, or more, sides may be described upon a given straight line similar to one given, and so on. PROP. XIX. THEOR. Similar triangles are to one another in the duplicate ratio of the homologous Let ABC, DEF be similar triangles, having the angle B equal to the angle E, and let AB be to BC, as DE to EF, so that the side BC is homologous to EF (def. 13. 5.) the triangle ABC has to the triangle DEF, the duplicate ratio of that which BC has to EF. sides. Take BG a third proportional to BC and EF (11. 6.), or such that AB: DE: BC: EF; but BC: EF :: EF: BG; therefore (11. 5.) AB: DE EF: BG; wherefore the sides of the triangles ABG, DEF, which are about the equal angles, are reciprocally propor tional; but triangles, which have the sides about two equal angles recipro cally proportional, are equal to one another (15.6.): therefore A D the triangle ABG is equal to the triangle DEF; and because that BC is to EF, as EF to BG; and that if three straight fines be proportionals, the first has to the third the duplicate ratio of that which it has to the second; BC therefore has to BG the duplicate ratio of that which BC has to EF. But as BC to BG, so is (1. 6.) the triangle ABC to the triangle ABG: therefore the triangle ABC has to the triangle ABG the duplicate ratio of that which BC has to EF: and the triangle ABG is equal to the triangle DEF; wherefore also the triangle ABC has to the triangle DEF the duplicate ratio of that which BC has to EF. B G CE F COR. From this, it is manifest, that if three straight lines be proportionals, as the first is to the third, so is any triangle upon the first to a similar, and similarly described triangle upon the second. > PROP. XX. THEOR. Similar polygons may be divided into the same number of similar triangles, having the same ratio to one another that the polygons have; and the polygons have to one another the duplicate ratio of that which their homologous sides have. Let ABCDE, FGHKL, be similar polygons, and let AB be the homologous side to FG: the polygons ABCDE, FGHKL, may be divided into the same number of similar triangles, whereof each has to each the same ratio which the polygons have; and the polygon ABCDE has to the polygon FGHKL a ratio duplicate of that which the side AB has to the side FG. Join BE, EC, GL, LH: and because the polygon ABCDE is similar to the polygon FGHKL, the angle BAE is equal to the angle GFL (def. 1. 6.), and BA: AE :: GF : FL (def. 1. 6.): wherefore, because the triangles ABE, FGL have an angle in one equal to an angle in the other, and their sides about these equal angles proportionals, the triangle ABE is equiangular (6. 6.), and therefore similar, to the triangle FGL (4. 5.): wherefore the angle ABE is equal to the angle FGL: and, because the polygons are similar, the whole angle ABC is equal (def. 1. 6.) to the whole angle FGH; therefore the remaining angle EBC is equal to the remaining angle LGH: now because the triangles ABE, FGL are similar, EB BA:: LG: GF; and also because the polygons are similar, AB : BC :: FG: GH (def. 1. 6.); therefore, ex æquali (22. 5.) EB: BC:: LG: GH, that is, the sides about the equal angles EBC, LGH are proportionals; therefore (6. 6.) the triangle EBC is equiangular to the triangle LGH, and similar to it (4. 6.). For the same reason, the triangle ECD is likewise similar to the triangle LHK; therefore the similar polygons ABCDE, FGHKL are divided into the same number of similar triangles. Also these triangles have, each to each, the same ratio which the polygons have to one another, the antecedents being ABE, EBC, ECD, and the consequents FGL, LGH, LHK: and the polygon ABCDE has to the polygon FGHKL the duplicate ratio of that which the side AE has to the homologous side FG. Because the triangle ABE is similar to the triangle FGL, ABE has to FGL the duplicate ratio (19. 6.) of that which the side BE has to the side GL: for the same reason, the triangle BEC has to GLH the duplicate ratio of that which BE has to GL: therefore, as the triangle ABE to the triangle FGL, so (11. 5.) is the triangle BEC to the triangle GLH. Again, because the triangle EBC is similar to the triangle LGH, EBC has to LGH the duplicate ratio of that which the side EC has to the side LH: for the same reason, the triangle ECD has to the triangle LHK, the duplicate ratio of that which EC has to LH: therefore, as the triangle EBC to the triangle LGH, so is (11. 5.) the triangle ECD to the triangle LHK: but it has been proved, that the triangle EBC is likewise to the triangle LGH, as the triangle ABE to the triangle FGL. Therefore, as the triangle ABE is to the triangle FGL, so is the triangle EBC to the triangle LGH, and the triangle ECD to the triangle LHK: and therefore, as one of the antecedents to one of the consequents, so are all the antecedents to all the consequents (12. 5.). Wherefore, as the triangle ABE to the tri angle FGL, so is the polygon ABCDE to the polygon FGHKL: but the triangle ABE has to the triangle FGL, the duplicate ratio of that which the side AB has to the homologous side FG. Therefore also the polygon ABCDE has to the polygon FGHKL the duplicate ratio of that which AB has to the homologous side FG. COR. 1. In like manner it may be proved, that similar figures of four sides, or of any number of sides, are one to another in the duplicate ratio of their homologous sides, and the same has already been proved of triangles: therefore, universally, similar rectilineal figures are to one another in the duplicate ratio of their homologous sides. COR. 2. And if to AB, FG, two of the homologous sides, a third proportional M be taken, AB has (def. 11. 5.) to M the duplicate ratio of that which AB has to FG; but the four-sided figure, or polygon, upon AB has to the four-sided figure, or polygon, upon FG likewise the duplicate ratio of that which AB has to FG: therefore, as AB is to M, so is the figure upon AB to the figure upon FG, which was also proved in triangles (Cor. 19. 6.). Therefore, universally, it is manifest, that if three straight lines be proportionals, as the first to the third, so is any rectilineal figure upon the first, to a similar, and similarly described rectilineal figure upon the second. COR. 3. Because all squares are similar figures, the ratio of any two squares to one another is the same with the duplicate ratio of their sides; and hence, also, any two similar rectilineal figures are to one another as the squares of their homologous sides. SCHOLIUM. If two polygons are composed of the same number of triangles similar, and similarly situated, those two polygons will be similar. For the similarity of the two triangles will give the angles EAB=LFG, ABE FGL, EBC=LGH: hence, ABC=FGH, likewise BCD=GHK, &c. Moreover, we shall have, EA: LF :: AB: FG:: EB: LG:: BC : GH, &c.; hence the two polygons have their angles equal and their sides proportional; consequently they are similar, Rectilineal figures which are similar to the same rectilineal figure, are also similar to one another. Let each of the rectilineal figures A, B be similar to the rectilineal figure C: The figure A is similar to the figure B. Because A is similar to C, they are equiangular, and also have their sides about the equal angles proportionals (def. 1. 6.). Again, because B is similar to C, they are equiangular, and have their sides about the equal angles proportionals (def. 1. 6.): therefore the figures A, B, are each of A B them equiangular to C, and have the sides about the equal angles of each of them, and of C, proportionals. Wherefore the rectilineal figures A and B are equiangular (1. Ax. 1.), and have their sides about the equal angles proportionals (11.5.). Therefore A is similar (def. 1. 6.) to B. 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, EF, GH be proportionals, viz. AB to 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 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 third proportional O; and because |