the triangles ABC, EAD have the same ratio to the triangle BAD wherefore the triangle ABC is equal (9. 5.) to the triangle EAD Therefore equal triangles, &c. Q. E. D. PROP. XVI. THEOR. If four straight lines be proportionals, the rectangle contained by the extremes is equal to the rectangle contained by the means; And if the rectangle contained by the extremes be equal to the rectangle contained by means, the four straight lines are proportionals. Let the four straight lines, AB, CD, E, F be proportionals, viz. as AB to CD, so E to F; the rectangle contained by AB, F is equal to the rectangle contained by CD, E. From the points A. C draw (11. 1) AG, CH at right angles to AB, CD; and make AG equal to F, and CH equal to E, and complete the parallelograms BG, DH Because AB CD::E: F; and since E=CH, and FAG, AB : CD (7. 5.)::CH: AG: therefore the sides of the parallelograms BG, DH about the equal angles are reciprocally proportional; but parallelograms which have their sides about equal angles reciprocally proportional, are equal to one another (14. 6.); therefore the parallelogram BG is equal to the parallelogram DH and the parallelogram BG is contained by the straight lines AB, F; because AG is equal to F; and F the parallelogram DH is contained by CD and E, because CH is equal to E therefore the rectangle contained by the straight lines AB, F is equal to that which is contained by CD and E. : : E Ꮐ B H D And if the rectangle contained by the straight lines AB, F be equal to that which is contained by CD, E; these four lines are proportionals, viz. AB is to CD, as E to F. The same construction being made, because the rectangle contained by the straight lines AB, F is equal to that which is contained by CD, E, and the rectangle BG is contained by AB, F, because AG is equal to F; and the rectangle DH, by CD, E, because CH is equal to E; therefore the parallelogram BG is equal to the parallelogram DH and they are equiangular: but the sides about the equal angles of equal parallelograms are reciprocally proportional (14.6.) wherefore AB CD::CH: AG, but CH=E, and AG=F, therefore AB : CD:: E F. Wherefore, if four, &c. Q. E. D. S 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 is (7. 5.) to B, as D to C: but if four straight lines be proportionals, the rectangle contained by the extremes is equal to that which is contained by the means (16. 6.): A therefore the rectangle AC the rectangle B.D; but the rectangle B.D is equal to the square of B, because B=D; therefore the rectangle A.C is equal to the square of B. B D C And if the rectangle contained by A, C be equal to the square of B; A: B::B: 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 (16. 6.): therefore A: B::D: C, but B=D; wherefore A: B:: B: C: Therefore, if three straight lines, &c. Q. E. D. PROP. XVIII. PROB. 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, make(23.1.) 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 AGB (32. 1.): wherefore the triangle FCD is equiangular to the triangle GAB: Again, at the points G, B in the straight line GB make (23. 1.) the angle BGH equal 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 : B 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: for the triangles GAB, FCD being equi. angular, BA: AG :: DC: CF (4. 6.); for the same reason, angular triangles BGH, DFE, GB: GH:: FD: FE; therefore, In the same manner, it may be proved, that AB BH: CD: DE. Also (4. 6.), GH: HB:: FE: ED. Wherefore, because the rectilineal figures ABHG, CDEF are equiangular, and have their sides about the equal angles proportionals, they are similar to one another (def. 1. 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 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.); herefore, 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 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 homelogous sides. 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. Take BG a third proportional to BC and EF (11. 6.), or such that BC EF EF: BG, and join GA. Then because AB: BC:: DE: EF, alternately (16. 5.), 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 recipro cally proportional: but triangles, which have the sides about two equal angles reciprocally proportional, are equal to one another (15.6.): therefore 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 lines be proportionals, the first has to the third the duplicate A 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. Therefore, similar triangles, &c. Q. E. D. COR. From this it is manifest, that if three straight lines be pre portionals, 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 homolgous 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 8 equiangular (6. 6.), and therefore similar, to the triangle FGL (4. 6.): 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.); there fore, 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, an similar to it (4. 6). For the same reason, the triangle ECD is like wise similar to the triangle LHK; therefore the similar polygons ABCDE, FGHKL are divided into the same number of similar triangles, |