PROP. XV. THEOR. EQUAL triangles which have one angle of the one equal to one angle of the other, have their fides about the equal angles reciprocally proportional. and triangles which have one angle in the one equal to one angle in the other, and their fides 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 fides about the equal angles of the triangles are reciprocally proportional; that is, CA is to AD, as EA to AB. Let the triangles be placed so that their fides CA, AD be in one straight line; wherefore also EA and AB are in one straight line; and join BD. Because the D Book VI. 2. 14. 1. triangle ABC is equal to the triangle ADE, and that ABD is another triangle; therefore as the triangle CАВ B is to the triangle BAD so is triangle A EA to AB; as therefore CA to AD, so is EA to AB d. wherefore d. 11. 5. the fides of the triangles ABC, ADE about the equal angles are reciprocally proportional. But let the fides of the triangles ABC, ADE about the equal angles be reciprocally proportional, viz. CA to AD, as EA to AB; the triangle ABC is equal to the triangle ADE. Having joined BD as before, because as CA to AD, fo is EA to AB; and as CA to AD, so is triangle BAC to triangle BAD and as EA to AB, so triangle EAD to triangle BAD; therefore d as triangle BAC to triangle BAD, so is triangle EAD to triangle BAD; that is, the triangles BAC, EAD have the fame ratio to the triangle BAD. wherefore the triangle ABC is equal to the tri- c. 9.5 angle ADE. Therefore equal triangles, &c. Q. F. D. I PROP. XVI. THEOR. F 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, viza 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 a 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 Fa and that E is equal to CH, and F to AG; AB is b to CD, as CH to AG. therefore the fides 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. and the parallelogram BG is con- DH is contained by CD and E, because CH is equal to E. A 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 conftruction being made, because the rectangle contained by the ftraight 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 parallelogram DH; and they are equiangular. but the fides about the equal angles of equal parallelograms are reciprocally pro- Book VI. portionals. wherefore as AB to CD, so is CH to AG; and CH is equal to E, and AG to F. as therefore AB is to CD, so E to F. c. 14. 6. Wherefore if four, &c. E. D. PROP. XVII. THEOR. F 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 Bis equal to D; A is a to B, as D to C. but if four straight lines a. 7. 5. be proportionals, the rec ABDC tangle contained by the extremes is equal to that which is contained by the means, therefore the rec tangle contained by A, C is equal to that contained by B, D. but the rectan gle contained by B, D is the square of RB; 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 fame 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 Di 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 propor tionals, therefore A is to B, as D. to. Ci but B is equal to D 1 Book VI. wherefore as A to B, so B to C. Therefore if three straight 4. 23. I. b. 32. 1. PON U a given straight line to describe a rectilineal figure fimilar, and similarly situated to a given rectilineal figure. Let AB be the given straight line, and CDEF the given rectilineal figure of four fides; it is required upon the given straight line AB to describe a rectilineal figure fimilar and fimilarly fituated to CDEF. Join DF, and at the points A, B in the straight line AB make a 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 AGBb. wherefore the triangle FCD is equiangular the remaining angle GHB, and the triangle FDE equiangular to the triangle GBH. then because the angle AGB is equal to the an gle 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 C. 4. 6. 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 likewife these 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, so, by reason of the equiangular triangles BGH, DFE, is FD to FE; d. 22. 5. therefore, ex aequalid, 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 ED. Wherefore because the rectilineal figures ABHG, CDEF are equiangular, and have their fides about Book VI. the equal angles proportionals, they are similar to one another. Next, Let it be required to defcribe upon a given straight line e. 1. Def. 6, AB, a rectilineal figure similar, and fimilarly situated to the recti lineal figure CDKEF of five fides. Join DE, and upon the given straight line AB describe the rectilineal figure ABHG fimilar and similarly situated to the quadrilateral figure CDEF, by the former cafe. 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 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; therefore ex c. 4. 6. aequalid, GH is to HL, as FE to EK. for the same reason, AB is d. 22. 5. to BL, 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 similar to one another. and in the fame manner a rectilineal figure of fix fides may be described upon a given straight line fimilar to one given, and fo on. Which was to be done. S PROP. XIX. THEOR. IMILAR triangles are to one another in the du. Let ABC, DEF be similar triangles having the angle B equal to Take BG a third proportional to BC, EF b, so that BC is to EF, b. 11, 6. as LF 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 as BC to LF, so 6. 16. 5. |