as DB to BE, † so is the parallelogram AB to †1.6. the parallelogram FE; and as GB to BF, so is the parallelogram BC to the parallelogram FE; therefore * as AB to FE, so BC to FE: there- *11.5, fore the parallelogram AB is equal * to the pa- * 9. 5, rallelogram BC. Therefore equal parallelograms, &c. Q. E. D. Equal triangles which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional: and triangles which have one angle in the one equal to one angle in the other, and their sides about the equal angles reciprocally proportional, are equal to one another. Let ABC, ADE be equal triangles, which have the angle BAC equal to the angle DAE: the sides about the equal angles of the triangles shall be reciprocally proportional; that is, CA shall be to AD, as EA to AB. Let the triangles be placed so B that their sides CA, AD be in one straight line; wherefore t also EA and AB are in one straight line; * and join BD. Because the triangle ABC is C equal to the triangle ADE, E *14, 1. and that ABD is another triangle; therefore as the triangle CAB, is to the triangle BAD,* so is *7.5. the triangle AED to the triangle DAB; but as the triangle CAB to the triangle BAD, so is the base CA to AD,* and as the triangle EAD to *1. 6. the triangle DAB, so is the base EA to AB;* *1, 6. therefore as CA to AD,* so is EA to AB: *11.5 wherefore the sides of the triangles ABC, ADE, about the equal angles are reciprocally proportional. Se the note to the last Proposition. * 1.6. * 1. 6. → 11.5. *9.5. *11. 1. +3. 1. +31. 1. *7.5. *14. 6. Next, let the sides of the triangles ABC, ADE about the equal angles be reciprocally proportional, viz. CA to AD, as EA to AB; the triangle ABC shall be equal to the triangle ADE. * Join BD as before: then because, as CA to AD, so is EA to AB; and as CA to AD, so is the triangle ABC to the triangle BAD; * and as EA to AB, so is the triangle EAD to the triangle BAD; therefore* as the triangle BAC to the triangle BAD, so is the triangle EAD to the triangle BAD; that is, the triangles BAC, EAD have the same ratio to the triangle BAD: wherefore the triangle ABC is equal to the triangle ADE. 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 the 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 shall be equal to the rectangle contained by CD, E. From the points A, C draw* 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, as AB to CD, so is E to F; and that E is equal to CH, and F to AG; AB is to CD as CH to 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; therefore the parallelogram BG is equal to the parallelogram DH: but the pa H rallelogram BG is contained by the straight lines A B C And if the rectangle contained by the straight lines AB, F be equal to that which is contained by CD, E; these four lines shall be proportional, viz. AB shall be 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 that 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 † 1AX. parallelogram DH; and they are equiangular : but the sides about the equal angles of equal parallelograms are reciprocally proportional: * *14. 6. wherefore, as AB to CD, so is CH to AG: but CH is equal to E, and AG to F; therefore as AB is to CD, † so is E-to F. Wherefore if four, +7.5. &c. Q. E. D. 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 rect angle contained by A, C shall be equal to the square of B. Take D equal to B: and because as A to B, *7.5. B to so B to C. and that B is equal to D; A is B, as D to C: but if four straight lines be proportionals, the rectargle A contained by the ex- D tremes is equal to that Cwhich is contained by 16.6. the means; * therefore the rectangle contained by A, C is equal to that D C A B contained 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 Â, 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 *16.6 lines are proportionals:* therefore A is to B, as D to C: but B is equal to D; wherefore, as A to B, so B to 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 *23. 1. line AB make the angle BAG equal to the G H and angle at C, and the angle ABG equal to the angle CDF; therefore the remaining angle CFD is equal to the remaining angle AGB:* therefore * 32. 1. the triangle FCD is equiangular to the triangle GAB: again, at the points G, B, in the straight lineGB,make* the angle BGH equal B K 23. 1. 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 +2 Ax. reason, the angle ABH is equal to the angle CDE: also the angle at A is equal † to the angle + Const. 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; because the triangles GAB, FCD being equiangular, BA is to AG, as DC to CF; and because * 4. 6. AG is to GB, as CF to FD; and as GB to GH, so, by reason of the equiangular triangles BGH, DFE, is FD to FE; therefore, ex æquali,* AG * 22. 5. 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 ED. Wherefore, *4. 6. because the rectilineal figures ABHG, CDEF are equiangular, and have their sides about the equal angles proportionals, they are similar to * 1 Def. 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. Join DE, and upon the given straight line AB describe the rectilineal figure ABHG similar, and similarly situated, to the quadrilateral 6. |
