1 quiangular: But the fides about the equal angles of equal pa- Book VI. rallelograms are reciprocally proportional: Wherefore as AB to CD, fo is CH to AG; and CH is equal to E, and AG to F: c 14. 6. As therefore AB is to CD, fo E to F. Wherefore, if four, &c. Q. E. D. PROP. XVII. THE OR. F three ftraight lines be proportionals, the rectangle contained by the extremes is equal to the fquare of the mean: And if the rectangle contained by the extremes be equal to the fquare of the mean, the three ftraight lines are proportionals. Let the three straight lines A, B, C be proportionals, viz. as A to B, fo B to C; the rectangle contained by A, C is equal to the fquare of B. Take D equal to B; and because as A to B, fo B to C, and that B is equal to D; A is to B, as D to C: But if four straight a 7.54 lines be proportionals, A the rectangle contain- But the rectangle con tained by B, D is the fquare of B; becaufe B is equal to D: Therefore the rectangle contained by A, C is equal to the fquare of B. And if the rectangle contained by A, C be equal to the fquare of B; A is to B, as B to C. The fame conftruction being made, because the rectangle contained by A, C is equal to the fquare of B, and the fquare 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 ftraight lines are proportionals: Therefore A is to B, as D to Book VI. C; but B is equal to D; wherefore as A to B, fo B to C. Therefore, if three ftraight lines, &c. Q. E. D. a 23.3. b 32. I UPON PON a given ftraight line to defcribe a rectilineal figure fimilar, and fimilarly fituated to a given rectilineal figure. Let AB be the given ftraight line, and CDEF the given rectilineal figure of four fides; it is required upon the given ftraight line AB to defcribe a rectilineal figure fimilar, and fimilarly fituated to CDEF. Join DF, and at the points A, B in the straight line AB 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 AGB: Wherefore the triangle FCD is equiangular to the triangle GAB: Again, at the points G, B in the ftraight line GB make the angle BGH equal to the G angle DFE, and the angle GBH equal to A FDE; therefore the H E F K C 4. 6. d 22. 5. 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 fame reason, the angle ABH is equal to the angle CDE; alfo 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 likewife thefe figures have their fides about the equal angles proportionals: Because 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, fo, by reafon of the equiangular triangles BGH, DFE, is FD to FE; therefore, ex aequali, AG is to GH, as CF to FE: In the fame manner it may be proved that AB is to BH as CD to DE: And GH is to HB, as FE to ED Wherefore, because the the rectilineal figures ABHG, CDEF are equiangular, and Book VI. have their fides about the equal angles proportionals, they are fimilar to one another *. Next, Let it be required to defcribe upon a given straight line AB, a rectilineal figure fimilar, and fimilarly fituated to the rectilineal figure CDKEF of five fides. e 1. def. 6. Join DE, and upon the given ftraight line AB defcribe the rectilineal figure ABHG fimilar, and fimilarly fituated to the quadrilateral figure CDEF, by the former cafe; and at the points B, H in the ftraight 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 fimilar, 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 fame reafon the angle ABL is equal to the angle CDK: Therefore the five fided figures AGHLB, CFEKD are equiangular: And because the figures AGHB, CFED are fimilar, GH is to HB, as FE to ED; and as HB to HL, fo is ED to EK; therefore, ex aequali d, c 4.6. GH is to HL, as FE to EK: For the fame reason, AB is to BL, d 22. si as CD to DK: And BL is to LH, as DK to KE, because the triangles BLH, DKE are equiangular: Therefore, because the five fided figures AGHLB, CFEKD are equiangular, and have their fides about the equal angles proportionals, they are fimilar to one another: And in the fame manner a rectilineal figure of fix or more fides may be described upon a given straight line fi milar to one given, and so on. Which was to be done. ST IMILAR triangles are to one another in the duplicate Let ABC, DEF be fimilar triangles having the angle B equal to the angle E, and let AB be to BC, as DE to EF, fo that the fide BC is homologous to EF: The triangle ABC has to the a 12. def. gs. triangle DEF, the duplicate ratio of that which BC has to EF. Take BG a third proportional to BC, EF, fo that BC is to b 11. 6. EF, as EF to BG, and join GA: Then, because as AB to BC, fo DE to EF; alternately, AB is to DE, as BC to EF: But c 16. 5. as Book VI. d 11.5. e 15. 6. g 1.6. as BC to EF, fo is EF to BG; therefore d as AB to DE, fo is EF to BG: Wherefore the fides of the triangles ABG, DEF which are about the equal angles are reciprocally proportional: But triangles which have the fides about two equal angles reciprocally proportional are equal to one ano- AA CE f 10. def. 5. faid f to have to the third the duplicate ratio of that which it has to the fecond; BC therefore has to BG the duplicate ratio of that which BC has to EF: But as BC to BG, fo 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 alfo the triangle ABC has to the triangle DEF the duplicate ratio of that which BC has to EF. Therefore imilar triangles, &c. Q. E. D. COR. From this it is manifeft, that if three ftraight lines be proportionals, as the firft is to the third, fo is any triangle upon the firft to a fimilar, and fimilarly defcribed triangle upon the fecond. SIMILAR polygons may be divided into the fame number of fimilar triangles, having the fame ratio to one another that the polygons have; and the polygons have to one another the duplicate ratio of that which their homologous fides have. Let ABCDE, FGHKL be fimilar polygons, and let AB be the homologous fide to FG: The polygons ABCDE, FGHKL may be divided into the fame number of fimilar triangles, whereof each to each has the fame ratio which the polygons have; and the polygon ABCDE has to the polygon FGHKL the duplicate ratio of that which the fide AB has to the fide FG. Join BE, EC, GL, LH: And because the polygon ABCDE is fimilar fimilar to the polygon FGHKL, the angle BAE is equal to the Book VI. angle GFL, and BA is to AE, as GF to FL: Wherefore, because the triangles ABE, FGL have an angle in one equal a 1. def. 6. to an angle in the other, and their fides about these equal angles proportionals, the triangle ABE is equiangular, and there- b 6. 6. fore fimilar to the triangle FGL; wherefore the angle ABE c 4.6. is equal to the angle FGL: And, becaufe the polygons are fimilar, the whole angle ABC is equal to the whole angle FGH; therefore the remaining angle EBC is equal to the remaining angle LGH: And becaufe the triangles ABE, FGL are fimilar, EB is to BA, as LG to GF; and alfo, because the polygons are fimilar, AB is to BC, as FG to GH; therefore, ex aequali, EB is to BC, as LG to GH; that is, the fides about & 22. 5. the equal angles EBC, LGH are proportionals; therefore b the triangle EBC is equiangular to the triangle LGH, and fi milar to it fimilar poly gons ABCDE, FGHKL are divided into the fame number of fimilar triangles. Also these triangles have, each to each, the fame ratio which the polygons have to one another, the antecedents being ABE, ÉBC, ECD, and the confequents FGL, LGH, LHK: And the polygon ABCDE has to the polygon FGHKL the duplicate ratio of that which the fide AB has to the homologous fide FG. e Because the triangle ABE is fimilar to the triangle FGL, ABE has to FGL the duplicate ratio of that which the fide e 19.6: BE has to the fide GL: For the fame 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, fo f is the f 11. 5. triangle BEC to the triangle GLH. Again, because the triangle EBC is fimilar to the triangle LGH, EBC has to LGH the duplicate ratio of that which the fide EC has to the fide LH: For the fame reason, the triangle ECD has to the triangle LHK, M |