* Take a straight line DE given in position and magnitude, and at the points D, E, make the angle EDF equal to the angle BAC, and the angle DEF equal to ABC; therefore the other angles EFD, BCA are equal, and each of the angles at the points A, B, C is given, wherefore each of those at the points D, E, F, is given: And because the straight line FD is drawn to the given point D in DE which is given in position, making the given angle EDF; therefore DF is given in position*; * 32 Dat. In like manner EF also is given in position; wherefore the point F is given: and the points D, E are given; therefore each of the straight lines DE, EF, FD is given in magnitude; wherefore the triangle DEF is given in species; and it is similar to the triangle ABC; which therefore is given in species. PROP. XLIV. * If one of the angles of a triangle be given, and if the sides about it have a given ratio to one another; the triangle is given in species. Let the triangle ABC have one of its angles BAC given, and let the sides BA, AC, about it have a given ratio to one another; the triangle ABC is given in species. Take a straight line DE given in position and magnitude, and at the point D, in the given straight line DE, make the angle EDF equal to the given angle BAC: wherefore the angle EDF is given; and because the straight line FD is drawn to the given point D in ED, which is given in position, making the given angle EDF; therefore FD is given in position *. And because the ratio of BA to AC is given, make the ratio of ED to DF the same with it, and join EF; and be D A A F * 29 Dat. 4 Dat. * 4.6. 41. 1 Def. 6. * 32 Dat. cause the ratio of ED to DF is given, and ED is given, therefore DF is given in magnitude: and it is given 2 Dat. also in position, and the point D is given, wherefore the point F is given *; and the points D, E are given, * 30 Dat. *29 Dat. 42 Dat. * 6.6. 42. See N. * 2 Dat. * 2 Dat. * 22. 5. 20.1. * wherefore DE, EF, FD are given in magnitude; and the triangle DEF is therefore given in species; and because the triangles ABC, DEF have one angle BAC equal to one angle EDF, and the sides about these angles proportionals; the triangles are similar, but the triangle DEF is given in species, and therefore also the triangle ABC. PROP. XLV. * If the sides of a triangle have to one another given ratios; the triangle is given in spécies. Let the sides of the triangle ABC have given ratios to one another, the triangle ABC is given in species. * B A C DEF H K Take a straight line D given in magnitude; and because the ratio of AB to BC is given, make the ratio of D to E the same with it; and D is given, therefore* E is given. And because the ratio of BC to CA is given, to this make the ratio of E to F the same; and E is given, and therefore * F. And because as AB to BC, so is D to E; by composition AB and BC together are to BC, as D and E to F: but as BC to CA, so is E to F; therefore, ex æquali *, as AB and BC are to CA, so are D and E to F, and AB and BC are greater than CA; therefore D and E are greater * than F. In the same manner, any two of the three D, E, F are greater than the third. Make* the triangle GHK, whose sides are equal to D, E, F, so that GH be equal to D, HK to E, and KG to F; and because D, E, F are, each of them, given, therefore GH, HK, KG are each of them given in magnitude; therefore the triangle GHK is * 42 Dat. given given in species: but as AB to BC, so is (D to E, that is) GH to HK; and as BC to CA, so is (E to F, that is) HK to KG; therefore, ex æquali, as AB to AC, so is GH to GK. Wherefore the triangle ABC is equiangular and similar to the triangle GHK; and the triangle GHK is given in species: therefore also the triangle ABC is given in species. * A. 5. 22.1. * 5. 6. COR. If a triangle be required to be made, the sides of 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. PROP. 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. 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 43. * 2 Dat. * 19. 1. A. 5. * * G * *6 Def. 32 Dat. 28 Dat. * 42 Dat. cumference 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 29 Dat. And because the triangles ABC, DEG, species*. 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 *7.6. 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. 44. See N. * 32. 1. * 43 Dat. * 32 Dat. * 2 Dat. * A. 5. * 6 Dat. 28 Dat. PROP. XLVII. If a triangle have one of its angles which is not a right angle 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 * B A A C 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 ratio of BA to AC is given, as BA to AC, so make ED to DG; and because the ratio of ED to DG is given, and ED is given, the straight line DG is given *, and BA is less than AC, therefore ED is less than DG. From the centre D, at the distance DG, describe the circle GF meeting EF in F, and join DF; and because the circle is given in position, as also E the straight line EF, the point F is given *; and the points D, E, are given; wherefore the straight lines DE, 29 Dat. EF, FD are given in magnitude, and the triangle DEF in species*. And the angle ACB is less therefore ACB is less 42 Dat. 18. 1. * 17. 1. because BA is less than AC, than the angle ABC, and than a right angle. In the same manner because ED is less 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 sides about the angles BAC, EDF, proportionals, and each of the other angles ACB, DEF, less than a right angle; the triangles ABC, DEF are similar, and DEF is given in species, *7. 6. wherefore the triangle ABC is also given in species. * B A Thirdly, let the given ratio be the ratio of a greater tó 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 AČB E * 32 Dat. D *7.6. GF * * is a right angle, and the triangle ABC is given in 43 Dat. species. 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 * * 8.5. A * 10.5. * A. 5. B D H |