( 43. ) * 2 Dat. 19. 1. • A. 5. *6 Def. ⚫ 28 Dat. 29 Dat. * 42 Dat. 7.6. which shall have the same ratios which three given straight lines D, E, F have to one another; it is necessary that every two of them be greater than the third. PROPOSITION XLVI. If the sides of a right-angled triangle about one of the acute angles, have a given ratio to one another, the triangle is given in species. Let the sides 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 species. * B E F 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, so 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, so is DE to EF; and AB is less than BC, therefore DE is less than EF. From the point D, draw DG at right angles to DE, and from the centre E, at the distance EF, describe a circle which shall meet DG in two points; let G be either of them, and join EG: therefore the circumference of the circle is given in position; and the straight line DG is given in position, because it is drawn to the given point D in DE given in position, in a given angle; therefore* the point G is given, and the points D, E are given, wherefore DE, EG, GD are given in magnitude, and the triangle DEG in species*. And because the triangles ABC, DEG have the angle BAC equal to the angle EDG, and the sides about the angles ABC, DEG proportionals, and each of the other angles BCA, EGD, less than a right angle; the triangle ABC is equiangular* and similar to the triangle DEG; but DEG is given in species; therefore the triangle ABC is given in species: and in the same manner, the triangle made by drawing a straight line from E to the other point in which the circle meets DG, is given in species. PROPOSITION XLVII. (44.) If a triangle have one of its angles which is not a right angle See N. given, and if the sides about another angle have a given ratio to one another, the triangle is given in species. Let the triangle ABC have one of its angles ABC a given, but not a right angle, and let the sides BA, AC, about another angle BAC, have a given ratio to one another: the triangle ABC is given in species. First, let the given ratio be the ratio of equality, that is, let the sides BA, AC, and consequently the angles ABC, ACB, be equal; and because the angle ABC is given, the angle ACB, and also the remaining* angle BAC is given; therefore the triangle ABC is given in species; and it is evident that in this case the given angle ABC must be acute. * B Next, let the given ratio be the ratio of a less to a greater, that is, let the side AB adjacent to the given angle be less than the side AC: take a straight line DE given in position and magnitude, and make the angle DEF equal to the given angle ABC: therefore EF * is given in position; and because the E B 82. 1. 43 Dat. 32 Dat. D 2 Dat. ratio of BA to AC is given, as BA to AC, • A. 5. 6 Dat. • 28 Dat. * 29 Dat. 42 Dat. 18. 1. * 17. 1. * 7. 6. 32 Dat. 7.6. * 43 Dat. * 8.5. * 10. 5. * A. 5. 7.6. ABC, DEF are similar, and DEF is given in species, wherefore the triangle ABC is also given in species. B Thirdly, let the given ratio be the ratio of a greater to a less, that is, let the side AB adjacent to the given angle be greater than AC; and, as in the last case, take a straight line DE given in position and magnitude, and make the angle DEF equal to the given angle ABC; therefore EF is given * in position also draw DG perpendicular to EF; therefore if the ratio of BA to AC be the same with the ratio of ED to the perpendicular DG, the triangles ABC, DEG are similar*, because the angles ABC, DEG are equal, and DGE is a right angle: therefore the angle ACB is a right angle, and the triangle ABC is given in species. * B EK But if, in this last case, the given ratio of BA to AC be not the same 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, because AC is greater than AM. Make as BA to AC, so 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 consequently DH is greater than DG; and because BA is greater than AC, ED is greater than DH. From the centre D, at the distance DH, describe the circle KHF which necessarily meets the straight line EF in two points, because DH is greater than DG, and less than DE. Let the circle meet EF in the points F, K, which are given, as was shewn in the preceding case; and DF, being joined, the triangles DEF, DEK are given in species, as was there shewn. From the centre A, at the distance AC, describe a circle meeting BC again in L: and if the angle ACB be less than a right angle, ALB must be greater than a right angle; and on the contrary. In the same manner, if the angle DEF be less than a right angle, DKE must be greater than one; and on the contrary. Let each of the angles ACB, DFE be either less or greater than a right angle: and because in the triangles ABC, DEF, the angles ABC. DEF are equal, and the sides BA, AC, and ED, DF, about two of the other angles proportionals, the triangle ABC is similar H DK to the triangle DEF. In the same manner the triangle ABL is similar to DEK. And the triangles DEF, DEK are given in species; therefore also the triangles ABC, ABL are given in species. And from this it is evident, that in this third case, there are always two triangles of a different species, to which the things mentioned as given in the proposition can agree. PROPOSITION XLVIII. If a triangle have one angle given, and if both the sides together about that angle have a given ratio to the remaining side, the triangle is given in species. Let the triangle ABC have the angle BAC given, and let the sides BA, AC together, about that angle, have a given ratio to BC: the triangle ABC is given in species. * (45.) 9. 1. • 3.6. 12. 5. Bisect the angle BAC by the straight line AD; therefore the angle BAD is given. And because as BA to AC, so is BD to DC, by permutation, as AB to BD, so is AC to CD; and as BA and AC together to BC, so 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 is given; therefore the triangle ABD is given in species, and 47 Dat. the angle ABD is therefore given; the angle BAC is also given, wherefore the triangle ABC is given in species *. A triangle which shall have the things that are mentioned 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 43 Dat. Bisect the angle EFG by the straight line FL, and by * 9. 1. 47th Prop. find a triangle of which EFL is one of the and in which the ratio of the sides about the angle te to FL is the same with the ratio of H to K: to do take FE given in position and magnitude, and draw pendicular to FL: then if the ratio of H to K be the th the ratio of FE to EL, produce EL, and let it meet : the triangle FEP is that which was to be found: for the given angle EFG; and because this angle is bi A A • 3.6. 18. 1. ( 46.) ⚫ 44 Dat. sected by FL, the sides EF, EP, together are to EP as F H K P E FE to 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 shewn 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 of 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 these add the angles NEF, EFN; therefore the angles NEF, EFG are less than the angles NEF, EFN, FNE, 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. PROPOSITION XLIX. If a triangle have 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 shewn in the preceding proposition. But the ratio of BA and AC together to BC is given; therefore, also, the ratio of AB to BD is given. And the angle ABD is given, wherefore * the triangle ABD is given in species; and consequently the an |