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 shall be 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 shall be proportionals: therefore A: B:: D: C; but B is equal to D; wherefore A: B :: fore if three straight lines, &c. *16.6. B: C. Q. E. D. There COR. The first is to the third as the square of the first to the square of the second, for A: C:: rectangle A⚫A: rectangle AC, but the rectangle A⚫C is equal to the square of B, therefore A: C: square of A: square of B. 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 first let CDEF the given rectilineal figure have four sides: it is required upon the given straight line AB to describe a rectilineal figure similar, and similarly situated, to CDEF. Draw DF, and at the points A, B in the straight line AB make the angle BAG equal to the angle at #23. 1. C, and the angle ABG equal to the angle CDF; therefore the remaining angle CFD is equal to the #32. 1. and remaining angle AGB: therefore the triangle FCD is equiangular to the triangle GAB: again, 3 Ax. #23. 1. angle BGH equal to G the angle DFE, and the angle GBH equal to FDE; therefore the remaining angle FED is equal to the remaining angle GHB, K 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 equalt 12 Ax. to the whole CFE; for a similar reason, the angle ABH is equal to the angle CDE: also the angle +Const. at A is equal to the angle at C, and the angle GHB to FED: therefore the rectilineal figure ABHG is equiangular to CDEF: likewise these figures have their sides about the equal angles proportionals; for the triangles GAB, FCD being equiangular, BA: AG:: DC: CF; and AG: GB:: CF: FD; and, by reason of the equiangular triangles BGH, DFE, GB: GH:: FD: FE; therefore* AG: GH:: CF: FE; in like manner it may be proved that AB: BH :: CD: DE; and GH: HB:: FE: ED.* Wherefore, because the rectilineal figures ABHG, CDEF are equiangular, and have their sides about the equal angles #1 Def. 6. proportionals, they are similar to one another. 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 of five sides. #11. 5. #4. 6. Draw 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 +32. 1. and equal to the remaining angle at L: and because 3 AX. the figures ABHG, CDEF are similar, the angle +1 Def. 6. 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 a like 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: HB :: FE: ED; but HB: HL:: ED: EK;* therefore, GH: HL:: FE: EK: for a like reason, *4. 6. #11. 5. #4.6. * AB: BL:: CD: DK: and BL: LH:: DK: 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. In a similar manner a rectilineal figure of six 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 as the squares of their homologous 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 may be homologous to EF: the triangle ABC shall be to the triangle DEF as the square of BC to the square of EF. *9 Def. 5. #11. 6. #10. 5. #2.5. Take BG a third proportional to BC, EF, so that BC may be to EF, as EF to BG, and join GA: then, because, AB: BC:: DE: EF; alternately, AB: DE:: BC: EF; but as BC: EF:: EF: BG; therefore,* AB: DE:: EF: BG: therefore 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 #15. 6. proportional, are equal* to one another; therefore the triangle ABG is equal to the triangle DEF: and because BC: EF:: EF: BG; and that if three straight lines be proportional, the A first is to the third as the square of the first to that of the #16., 6. Cor. second; therefore BC is to BG as the square of BC #1. 6. is to the square of EF: but as BC to BG, so is the triangle ABC to the triangle ABG; therefore the triangle ABC is to the triangle ABG as the square of BC to the square of EF: but the triangle ABG is equal to the triangle DEF; therefore also the triangle ABC is to the triangle DEF as the square of BC to the square of 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. Otherwise : Upon BC, EF, but on opposite sides to the triangles, construct squares BG, EH. Through A and D draw parallels to BC, EF respectively, and prolong the sides of the squares to meet them. The figures BK, EL thus formed are rectangles. Because MK, BC are parallels, and AB meets them, the angle MAB is equal to the angle ABC.* #29. 1. For similar reasons, the angle NDE is equal to the angle DEF. But the angles ABC, DEF are equal; there #29. 1. fore the angles MAB, NDE are equal. Moreover the angles M and N are right angles; therefore the angles MBA, NED are equal; that is the triangles MBA, NED are equiangular, and therefore similar. Consequently,* #4. 6. or, since BC is equal to BP and EF to EQ, #1.6. *2.5. MB: BP: NE: EQ. But MB is to BP as MC to BG,* and NE is to MC BG: NF: EH, #10. 5. and alternately* MC NF: BG: EH. But the parallelograms MC, NF being doubles of the triangles are to each other as the triangles.* Therefore the triangles ABC, DEF are to each other as the squares on the homologous sides BC, EF. Q. E. D. #5. 5. Cor. 1. PROP. XX. THEOR. Similar polygons may be divided into the same number of similar triangles, which are to each other as the polygons themselves; and the polygons are to one another as the squares of their homologous sides. Let ABCDE, FGHKL be similar polygons, and let AB be the homologous side to FG: the polygons may be divided into the same number of similar triangles, whereof each shall be to each as the polygons are to each other; and the polygon ABCDE shall be to the polygon FGHKL as the square of the side AB to the square of the side FG. Draw EB, EC, LG, LH, &c., and because the polygon ABCDE is similar to the polygon FGHKL, the angle BAE is equal to the angle GFL, and BA: AE :: GF: FL; therefore, because the triangles ABE, FGL *1 Def. 6. |