IF 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 a 44. dat. BD is given. And the angle ABD is given, wherefore a 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 is b. 43. dat. given in species b. B HK F A C D E M L G A triangle which shall have the things mentioned in the proposition to be given, may be thus found. Let EFG be the given angle, and the ratio of H to K the given ratio; and by prop. 44 find the triangle EFL, 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. IF 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 spe cies, let AD be drawn to the base BC in a given angle ADB; the ratio of AD to BC is given. Because the triangle ABC is given in species, the angle ABD is given, and the angle ADB is given, therefore the triangle ABD is given a in species; wherefore the ratio of AD to AB is given. And the ratio of AB to BC is given; and therefore the ratio of AD to BC is given. PROP. LI. B A 76. a 43. dat. D C b 9. dat. 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 B A 47. a 3. def. b 44. dat E angle BAE is given a, and the ratio of BA C D wherefore the tri- c 9. dạt. 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; the angle DAB, E that is, the angle AEC C tio of BA to AD; But therefore the ratio of b EA to AD is given, and the triangle ACB PROBLEM. To find the ratio of two triangles ABC, ABD given in species, and which are described upon the same straight line AB. e 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 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; therefore the angles ACE, AEC are equal to FHL, FLH, each to each, and the triangle AEC equiangular to the triangle FLHI. 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 aequali, 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. IF two rectilineal figures given in species be de. See Note 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. Join AC, AD, AF: each of the triangles AED, ADC, ACB, AGF, ABF is given a in species. And because the triangles a 51, dat ADE, ADC given in species are de D scribed upon the same straight line AD, the ratio of EAD to DAC is E given b; and, by composition, the C b 52. dat. c 7. dat. A B G F ratio of EACD to DAC is given. And the ratio of DAC to CAB is given b, because they are described upon the same straight line AC; therefore the ratio of EACD to ACB is given ; and, by composition, the K L M N ratio of ABCDE to ABC is given. H|~|~||~o d9. dat. In the 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 4. 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 50. a 9. dat. b2 Cor. 20.6. to KL the same with it. Find also the ratio of the triangle ACD G E A H K L M N C B F Because the triangle EAD is to the triangle DAC as the straight line HK to KL; and as the triangle DAC to CAB, so is the straight line KL to LM; therefore, by using composition as often as the number of triangles requires, the rectilineal ABCDE is to the triangle ABC, as the straight line HM to ML. In like manner, because the triangle GAF is to FAB, as ON to NM, by composition, the rectilineal ABFG is to the triangle ABF, as MO to NM; and, by inversion, as ABF to ABFG, so is NM to MO. And the triangle ABC is to ABF, as LM to MN. Wherefore, because as ABCDE to ABC, so is HM to ML; and as ABC to ABF, so is LM to MN; and as ABF to ABFG, so is MN to MO: ex æquali, as the rectilineal ABCDE to ABFG, so is the straight line HM to MO. PROP. LIV. IF two straight lines have a given ratio to one another, the similar rectilineal figures described upon them similarly, shall have a given ratio to one another. Let the straight lines AB, CD have a given ratio to one another, and let the similar and similarly placed rectilineal figures E, F be described upon them; the ratio of E to F is given. To AB, CD, let G be a third proportional: therefore, as AB to CD, so is CD to G. And the ratio of AB. to CD is given, wherefore the ratio of CD O to G is given; and consequently the ratio of AB to G is also given a. But as AB to G, so is the figure E to the figure b F. There fore the ratio of E to F is given. A E G F BC D H K L |
