ed in the proposition 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 ratio which the two sides about the angle EFG must have to the third side of the triangle; therefore because two sides of a triangle are greater than the third side, the ratio of H to K must be the ratio of a greater to a less. Bisect a the angle EFG by the straight 9. 1. line FL, and by the 47th proposition find a triangle of, which EFL is one of the angles, and in which the ratio of the sides about the angle opposite to FL is the same with the ratio of H to K: To do which, take FE given in position and magnitude, and draw EL perpendicular to FL: Then if the ratio of H to K be the same 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 be F H K G P cause this angle is bisected by FL, the sides EF, FP, together are to EP, asb FE to EL, that is, as H to K. E 3. 6. But if the ratio of H to K be not the same with the ratio of FE to EL, it must be less than it, as was shown in Prop. 47, and in this case there are two triangles, each of which has the given angle EFL, and the ratio of the sides about the angle opposite to FL the same with the ratio of H to K. By Prop. 47, find these triangles EFM, EFN, each of which has the angle EFL for one of its angles, and the ratio the side FE to EM or EN the same with the ratio of H to K; and let the angle EMF be greater, and ENF less, than a right angle. And because H is greater than K, EF is greater than EN, and therefore the angle EFN, that is, the angle NFG, is less than the angle ENF. To each of '18. 1. these add the angles NEF, EFN; therefore the angles NEF, EFG, are less than the angles NEF, EFN, FNË, that is, than two right angles; therefore the straight lines EN, FG, must 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 proposition: For each of them has the given angle EFG; and because this angle is bisected by the straight line FMN, the sides EF, FG, together have to EG the third side the ratio of FE to EM, that is, of H to K. In like manner, the sides EF, FO, together have to EO the ratio which H has to K. Ir a triangle has one angle given, and if the sides about another angle both together have a given ratio to the third side; the triangle is given in species. Let the triangle ABC have one angle ABC given, and let the two sides BA, AC, about another angle BAC have a given ratio to BC; the triangle ABC is given in species. Suppose the angle BAC to be bisected by the straight line AD; BA and AC together are to BC, as AB to BD, as was shown in the preceding proposition. But the ratio of BA and AC together to BC is given; therefore also the ratio of AB to BDis given. And the angle ABD is given, ⚫ 44 Dat. wherefore the triangle ABD is given in species; and consequently the angle BAD, and its double the angle BAC are given: And the angle ABC is given. Therefore the triangle ABC b 43 Dat. is given in species b. F E M G A triangle which shall have the things mentioned in the proposition B to be given, may be thus found. Let EFG be the given angle, and the ratio of H to K the given ratio; and Hby Prop. 44. find the triangle EFL, K which has the angle EFG for one of its angles, and the ratio of the sides EF, FL, about this angle the same with the ratio of H to K; and make the angle LEM equal to the angle FEL. And because the ratio of H to K is the ratio which two sides of a triangle have to the third, H must be greater than K: And because EF is to FL, as H to K; therefore EF is greater than FL, and the angle FEL, that is, LEM, is therefore less than the angle ELF. Wherefore the angles LFE, FEM, are less than two right angles, as was shown in the foregoing proposition, and the straight lines FL, EM, must meet if produced; let them meet in G, EFG is the triangle which was to be found: for EFG is one of its angles, and because the angle FEG is bisected by EL, the two sides FE, EG, together have to the third side FG the ratio of EF to FL, that is, the given ratio of H to K. PROP. L. Is from the vertex of a triangle given in species, a straight line be drawn to the base in a given angle; it shall have a given ratio to the base. From the vertex A of the triangle ABC which is given in species, let AD be drawn to the base BC in a given angle ADB; the ratio of AD to BC is given. 76. 43 Dat. Because the triangle ABC is given in species, the angle ABD is given, and the angle ADB is given, therefore the triangle ABD is givena in species; wherefore the ratio of AD to AB is given. And the ratio of AB to BC is given; and therefore b the ratio ↳ 9 Dat. of AD to BC is given. B PROP. LI RECTILINEAL figures given in species, are divided into triangles which are given in species. Let the rectilineal figure ABCDE be given in species: ABCDE may be divided into triangles given in species. Join BE, BD: and because ABCDE is given in species, the angle BAE is givena, and the ratio of BA to AE is given; wherefore the triangle BAE is given in species, and the angle AEB is therefore givena. B But the whole angle AED is given, and therefore the remaining angle BED is given, and the ratio of AE to E Ꭰ . 47. a 3 Def. b 44 Dat. с EB is given, as also the ratio of AE to ED; therefore the ratio of BE to ED is given. And the angle BED is 9 Dat. given, wherefore the triangle BED is given in species. In the same manner, the triangle BDC is given in species: Therefore rectilineal figures which are given in species are divided into triangles given in species. If two triangles given in species be described upon the same straight line; they shall have a given ratio to one another. Let the triangles ABC, ABD, given in species be described upon the same straight line AB; the ratio of the triangle ABC to the triangle ABD is given. Through the point C, draw CE parallel to AB, and let it meet DA produced in E, and join BE. Because the triangle ABC is given in species, the angle BAC, that is, the angle ACE, is given; and because the triangle ABD is given in species, theE angle DAB, that is, 3 Def. AC is givena, and the D C L H B F K ratio of CA to AB is given, as also the ratio of BA to AD; 9 Dat. therefore the ratio of EA to AD is given, and the triangle 37. 1. ACB is equal to the triangle AEB, and as the triangle d 1. 6. AEB, or ACB, is to the triangle ADB, so isd the straight line EA to AD: But the ratio of EA to AD is given; therefore the ratio of the triangle ACB to the triangle ADB is given. PROBLEM. To find the ratio of two triangles ABC, ABD, given in species, and which are described upon the same straight line AB. Take a straight line FG given in position and magnitude, and because the angles of the triangles ABC, ABD, are given, at the points F, G, of the straight line FG, make 23. 1. the angles GFH, GFK, equal to the angles BAC, BAD ; and the angles FGH, FGK, equal to the angles ABC,ABD, each to each. Therefore the triangles ABC, ABD, are equiangular to the triangles FGH, FGK, each to each. Through the point H draw HL parallel to FG, meeting KF produced in L. And because the angles BAC, BAD, are equal to the angles GFH, GFK, each to each; there fore the angles ACE, AEC, are equal to FHL, FLH, each to each, and the triangle AEC equiangular to the triangle FLH. Therefore as EA to AC, so is LF to FH, and as CA to AB, so HF to FG; and as BA to AD, so is GF to FK; wherefore, ex æquali, as EA to AD, so is LF to FK. But as was shown, the triangle ABC is to the triangle ABD, as the straight line EA to AD, that is, as LF to FK. The ratio therefore of LF to FK has been found, which is the same with the ratio of the triangle ABC to the triangle ABD. Ir two rectilineal figures given in species be de- See N. scribed upon the same straight line; they shall have a given ratio to one another. Let any two rectilineal figures ABCDE, ABFG, which are given in species, be described upon the same straight line AB; the ratio of them to one another is given. E a b 52 Dat. Join, AC, AD, AF; each of the triangles AED, ADC, ACB, AGF, ABF, is given in species. And because the 51 Dat. triangles ADE, ADC, given in species, are described upon the same straight line AD, the ratio of EAD to DAC is givenb; and by composition, the ratio of EACD to DAC is given. And the ratio DAC to CAB, is given, because they are described upon the same. straight line AC; therefore the ratio of EACD to ACB is givend;" and, by composition, the ratio of, G. A C 7 Dat. B ABCDE to ABC is given. In the HKL MN O . same manner, the ratio of ABFG to ABF is given. But the ratio of the triangle ABC to the triangle ABF is given; wherefore because the ratio of ABCDE to ABC is given, as also the ratio of ABC to ABF, and the ratio of ABF to ABFG; the ratio of the rectilineal ABCDE to the rectilineal ABFG is given". PROBLEM. To find the ratio of two rectilineal figures given in species, and described upon the same straight line. Let ABCDE, ABFG, be two rectilineal figures given in species, and described upon the same straight line AB, and join AC, AD, AF. Take a straight line HK given in position and magnitude, and by the 52d dat. find the ratio of the triangle ADE to the triangle ADC, and make the ratio of HK to KL the same with it. Find also the ratio of the d 9 Dat. |