DH; and they are equiangular: But the sides about the Boox VI. equal angles of equal parallelograms are reciprocally proportional: Wherefore, as AB to CD, so is CH to AG; 14. 6. and CH is equal to E, and AG to F: as therefore AB is to CD, so is E to F. Wherefore, if four, &c. Q.E.D. C PROP. XVII. THEOR. If three straight lines be proportionals, the rectangle contained by the extremes is equal to the square of the mean; and if the rectangle contained by the extremes be equal to the square of the mean, the three straight lines are proportionals. Let the three straight lines A, B, C be proportionals, viz. as A to B, so B to C; the rectangle contained by A, C is equal to the square of B. Take D equal to B; and because as A to B, so B to C, and that B is equal to D: A isa to B, as D to C: But if17. 5. four straight lines A be proportionals, the tained by B, D: But the rectangle contained by B, D is the square of B; because B is equal to D: Therefore the rectangle contained by A, C is equal to the square of B. And if the rectangle contained by A, C be equal to the square of B; A is to B, as B to C. The same construction being made, because the rectangle contained by A, C is equal to the square of B, and the square of B is equal to the rectangle contained by B, D, because B is equal to D; therefore the rectangle contained by A, C is equal to that contained by B, D; but if the rectangle contained by the extremes be equal to that contained by the means, the four straight lines are proportionals: Therefore A is to B, as D to C; but B is equal b 16. 6. Book VI. to D; wherefore, as A to B, so B to C: Therefore if three straight lines, &c. Q. E.D. PROP. XVIII. THEOR. See N. UPON a given straight line to describe a rectilineal figure similar, and similarly situated, to a given rectilineal figure. Let AB be the given straight line, and CDEF the given rectilineal figure of four sides; it is required upon the given straight line AB to describe a rectilineal figure similar, and similarly situated, to CDEF. Join DF, and at the points A, B in the straight line AB -23. 1. make the angle BAG equal to the angle at C, and the angle ABG equal to the angle CDF; therefore the remaining angle CFD is equal to the remaining angle 82. 1. AGB. Where fore the triangle FCD is equian- gle GAB: Again, H F K at the points G, B in the straight line GB,makea thean- A gle BGH equal to the angle DFE, and the angle GBH equal to FDE; therefore the remaining angle FED is 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, and 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: Therefore the rectilineal figure ABHG is equiangular to CDEF: But likewise these figures have their sides about the equal angles proportionals; because 4. 6. the triangles GAB, FCD being equiangular, BA is to AG, as DC to CF; and because AG is to GB, as CF to FD; and as GB to GH, so, by reason of the equiangular tri22. 5. angles BGH, DFE, is FD to FE; therefore, ex æquali4, AG is to GH, as CF to FE: In the same manner it may be proved that AB is to BH, as CD to DE; and GH is to HB, as FE to EDe. Wherefore, because the rectilineal Boos VI. figures ABHG, CDEF are equiangular, and have their sides about the equal angles proportionals, they are similar 4. 6. to one another f. 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 an, gle 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 EKs; therefore, ex æquali, 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, ass DK to KE, because the triangles BLH, DKE are equiangular: Therefore be cause the five-sided figures AGHLB, CFEKD are equi angular, and have their sides about the equal angles proportionals, they are similar to one another; and in the same manner a rectilineal figure of six or more sides may be described upon a given straight line similar to one given, and so on. Which was to be done. PROP. XIX. THEOR. SIMILAR triangles are to one another in the duplicate ratio of their homologous sides. f 1 Def. 6. h 4. 6. 22. 5. 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: the triangle 12 Def. 5. ABC has to the triangle DEF, the duplicate ratio of that which BC has to EF. b Take BG a third proportional to BC, EF, so that BC 11. 6. is to EF, as EF to BG, and join GA: Then, because, as AB to BC, so DE to EF; alternately, AB is to DE, as € 16. 5. d 11. 5. A BOOK VI. BC to EF: But as BC to EF, so is EF to BG; therefore as AB to DE, so is EF to BG. Wherefore the sides of the triangles ABG, DEF, which are about the equal angles, are reciprocally proportional: But triangles which have the sides about two equal angles reciprocally proportional are equal 15. 6. to one anothere: Therefore the triangle ABG is equal to the triangle DEF: And B G C D A because as BC is to EF, so EF to BG; and that if three £10 Def. 5. straight lines be proportional, the first is said to have to the third the duplicate ratio of that which it has to the second; BC therefore has to BG the duplicate ratio of that $1.6 which BC has to EF: But as BC to BG, so is 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: But 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. Therefore similar triangles, &c. Q. E.D. 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 to each has the same ratio which the polygons have; and the polygon ABCDE has to the polygon FGHKL the duplicate ratio of that which the side AB has to the side FG.. Join BE, EC, GL, LH: And because the polygon b C a 1 Def. 6. ABCDE is similar to the polygon FGHKL, the angle Book VI. BAE is equal to the angle GFL, and BA is to AE, as GF to FL: 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 equiangularb, and therefore similar, to the triangle ↳ 6. 6. FGL; wherefore the angle ABE is equal to the angle 4. 6. FGL: And, because the polygons are similar, the whole angle ABC is equala to the whole angle FGH; therefore the remaining angle EBC is equal to the remaining angle LGH: And because the triangles ABE, FGL are similar, EB is to BA, as LG to GF; and also, because the polygons are similar, AB is to BC, as FG to GHa; therefore, ex æquali, EB is to BC, as LG to GH: that is, the sides a 22. 5. about the equal angles EBC, LGH are proportionals; therefored the triangle EBC is equiangular to the triangle LGH, and similar to it. For the same reason, the triangle ECD likewise is 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 AB has to the homologous side FG. Because the triangle ABE is similar to the triangle FGL, ABE has to FGL, the duplicate ratio of that which the 19. 6. 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, sof is the triangle BEC to the triangle GLH. 11. 5. 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, triangle ECD has to the triangle LHK the duplicate ratio the |