PROP. XLVI. 43. IF the fides of a right angled triangle about one of the acute angles have a given ratio to one another; the triangle is given in fpecies. Let the fides AB, BC about the acute angle ABC of the triangle ABC which has a right angle at A, have a given ratio to one another; the triangle ABC is given in fpecies. Take a straight line DE given in position and magnitude; and because the ratio of AB to BC is given, make as AB to BC, fo DE to EF; and because DE has a given ratio to EF, and DE is given, therefore EF is given. and because as AB to BC, fo is DE to EF, a. 2. Dat. and AB is less than BC, therefore DE is lefs than EF. from the a b point D draw DG at right b. 19 I. c. A. 5. EG; therefore the circumference of the circle is given in pofition. d 6. Def. and the straight line DG is given in pofition, because it is drawn c. 32. Dat. to the given point D in DE given in pofition, in a given angle. therefore f the point G is given. and the points D, E are given, f. 28. Dat. wherefore DE, EG, GD are given & in magnitude, and the triangle g. 29. Dat. DEG in fpecies. and becaufe the triangles ABC, DEG have the h. 42. Dat. angle BAC equal to the angle EDG, and the fides about the angles ABC, DEG proportionals, and each of the other angles BCA, EGD less than a right angle; the triangle ABC is equiangular i and fimi- i. 7. 6. lar to the triangle DEG. but DEG is given in species, therefore the triangle ABC is given in fpecies. and in the fame manner, the triangle made by drawing a straight line from E to the other point in which the circle meets DG is given in fpecies. 44. See N. 1 IF Fa triangle has one of its angles which is not a right angle given, and if the fides about another angle have a given ratio to one another; the triangle is given in fpecies. Let the triangle ABC have one of its angles ABC a given, but not a right angle, and let the fides BA, AC about another angle BAC have a given ratio to one another; the triangle ABC is given in fpecies. First, Let the given ratio be the ratio of equality, that is, let the fides BA, AC and confequently the angles ABC, ACB be equal. and because the angle ABC is given, the angle ACB, a. 32. 1. and also the remaining angle BAC is given. b. 43. Dat. therefore the triangle ABC is given in fpecies. B and it is evident that in this cafe the given angle ABC must be acute. a A Next, Let the given ratio be the ratio of a lefs to a greater, that is, let the fide AB adjacent to the given angle be lefs than the fide AC. take a straight line DE given in pofition and magnitude, and make the angle DEF equal to the given angle ABC; therefore EF c. 32. Dat. is given in pofition, and becaufe d. 2. Dat. c. A. s. the ratio of BA to AC is given, as and because the ratio of ED to DG is given, and ED is given, the ftraight line DG is given 4. and BA is lefs than AC, therefore ED is less than DG. from the center D, at the diftance DG defcribe the circle GF meeting EF in F, and join DF. and f. 6. Def. because the circle is given f in pofition, as alfo the ftraight line EF, the G E A g. 18. Dat. point F is given . and the points D, E are given, wherefore the h. 29. Dat. ftraight lines DE, EF, FD are given h in magnitude, and the trii. 42. Dat. angle DEF in fpecies 1. and because BA is lefs than AC, the angle k. 18. I. ACB is lefsk than the angle ABC, and therefore ACB is lefs than a right I. 17. 1. a right angle. in the fame manner, because ED is lefs than DG or DF, the angle DFE is less than a right angle. and because the triangles ABC, DEF have the angle ABC equal to the angle DEF, and the fides about the angles BAC, EDF proportionals, and each of the other angles ACB, DFE lefs than a right angle; the triangles ABC, DEF are fimilar. and DEF is given in fpecies, wherefore m. 7. 6. the triangle ABC is also given in fpecies. A Thirdly, Let the given ratio be the ratio of a greater to a lefs, that is, let the fide AB adjacent to the given angle be greater than AC. and, as in the last case, take a straight line DE given in pofition and magnitude, and make the angle DEF equal to the given angle ABC; therefore EF is given in pofition. alfo draw DG perpendicular to EF; therefore if the ratio of BA to AC be the fame with the ratio of ED to the perpendicular DG, the triangles ABC, DEG are fimilar, becaufe the angles ABC, DEG are equal, and DGE is a right B 4 D c. 32. Dat. E G F angle. therefore the angle ACB is a right angle, and the triangle ABC is given bin fpecies. b. 43. Dat. A But if, in this laft cafe, the given ratio of BA to AC be not the fame with the ratio of ED to DG, that is, with the ratio of BA to the perpendicular AM drawn from A to BC; the ratio of BA to AC must be less than the ratio of BA to AM, becaufe AC is o. 8. 5. greater than AM. make as BA to AC, fo ED to DH; therefore the ratio of ED to DH is less than the ratio of (BA to AM, that is than the ratio of) ED to DG; and confequently DH is greater p than DG; and B because BA is greater than AC, ED is greater than DH. from the center D, at the diftance DH, defcribe the circle KHF which neceffarily meets the ftraight line EF in two points, because DH is greater than E K DG, and less than DE. let the circle meet L C P. 10. S. D e. A. s. F H EF in the points F, K which are given, as was fhewn in the preceeding cafe; and, DF, DK being joined, the triangles DEF, DEK are given in fpecies, as was there fhewn. from the center A, at the distance AC defcribe a circle meeting BC again in L. and if the angle 394 A angle ACB be lefs than a right angle, ALB must be greater than a right angle; and on the contrary. in the fame manner, if the angle DFE be lefs than a right angle, DKE muft be greater than one; and on the contrary. let each of the angles ACB, DFE be either lefs or greater than a right angle; and because in the triangles ABC, DEF the angles ABC, DEF are equal, and the fides BA, AC, and ED, DF B about two of the other angles proportiom. 7. 6. nals, the triangle ABC is fimilar to the triangle DEF. in the fame manner, the tri angle ABL is fimilar to DEK. and the tri L M C D angles DEF, DEK are given in fpecies, E K F 45. a. 9. I. given in fpecies. and from this it is evident, that, in this third cafe, there are always two triangles of a different fpecies to which the things mentioned as given in the Propofition can agree. IF PROP. XLVIII. Fa triangle has one angle given, and if both the fides together about that angle have a given ratio to the remaining fide; the triangle is given in fpecies. Let the triangle ABC have the angle BAC given, and let the fides BA, AC together about that angle have a given ratio to BC; the triangle ABC is given in fpecies. a Bifect the angle BAC by the ftraight line AD; therefore the b. 3. 6. angle BAD is given. and because as BA to AC, fo is b BD to DC, by permutation, as AB to BD, fo is AC to CD; and as BA and AC together to BC, fo C c. 12. 5. is AB to BD. but the ratio of BA and AC together to BC is given, wherefore the ratio of AB to BD is given; and the angle BAD B d. 47. Dat. is given, therefore the triangle ABD is gi A D ven in fpecies. and the angle ABD is therefore given; the angle e. 43. Dat. BAC is alfo given, wherefore the triangle ABC is given in fpecies. A triangle which fhall have the things that are mentioned in the Propofition to be given, can be found in the following manner. let EFG be the given angle, and let the ratio of H to K be the given 1 ratio which the two fides about the angle EFG must have to the third fide of the triangle. therefore because two fides of a triangle are greater than the third fide, the ratio of H to K must be the ratio of a greater to a lefs. bifect the angle EFG by the ftraight line a. 9. 1. FL, and by the 47 th Propofition find a triangle of which EFL is one of the angles, and in which the ratio of the fides about the angle oppofite to FL is the fame with the ratio of H to K; to do which, take FE given in pofition and magnitude, and draw EL perpendicular to FL. then, if the ratio of H to K be the fame with the ratio of FE to EL, produce EL and let it meet FG in P; the triangle FEP is that which was to be found. for it has the given angle EFG, and because this angle is bifected by FL, the fides EF, FP together are to EP, asb FE to EL, that is as H to K. H K b. 3. 6. G this cafe there are two triangles cach of which has the given angle EFL, and the ratio of the fides about the angle oppofite to FL the fame with the ratio of H to K. by Prop. 47. find thefe triangles EFM, EFN each of which has the angle EFL for one of its angles, and the ratio of the fide FE to EM or EN the fame with the ratio of H to K; and let the angle EMF be greater, and ENF lefs than a right angle. and becaufe H is greater than K, EF is greater than EN, and therefore the angle EFN, that is the angle NFG, is lefs f f. 18. 1. than the angle ENF. to each of thefe add the angles NEF, EFN; therefore the angles NEF, EFG are lefs than the angles NEF, EFN, FNE, that is than two right angles; therefore the ftraight lines EN, FG muft-meet together when produced; let them meet in O, and produce EM to G. each of the triangles EFG, EFO has the things mentioned to be given in the Propofition. for each of them has the given angle EFG, and because this angle is bifected by the ftraight line FMN, the fides EF, FG together have to EG the third fide the ratio of FE to EM, that is of H to K. in like manner, the fides EF, FO together have to EO the ratio which H has to K. PROP. |