81. and because the ratios of H to K, and of K to L are given, the ratio of H to L is given (9. dat.); because, therefore, as AEB to FD, so is H to K; and as FD to AG, so is K to L; ex æquali, as AEB 10 AG, so is H to L; therefore the ratio of AEB to AG is given ; and the figure AEB is given in species, and to its side AB the parallelogram AG is applied in the given angle ABG; therefore by the 69th dat. a parallelogram may be found similar to AG : let this be the parallelogram MN; MN also is similar to FD; for, by the construction, MN is similar to AG, and AG is similar to FD; therefore the parallelogram FD is similar to MN. PROP. LXXI. If the extremes of three proportional straight lines have given ratios to the extremes of other three proportional straight lines; the means shall also have a given ratio to one another: and if one extreme have a given ratio to one extreme, and the mean to the mean ; likewise the other extreme shall have to the other a given ratio. Let A, B, C be three proportional straight lines, and D, E, F three other; and let the ratios of A to D, and of C to F be given; then the ratio of B to E is also given. Because the ratio of A to D, as also of C to F is given, the ratio of the rectangle A, C to the rectangle D, F is given (67. dat.); but the square of B is equal (17. 6.) to the rectangle A, C; and the square of E to the rectangle (17. 6.) D, F; therefore the ratio of the square of B to the square of Eis given ; wherefore (58. dat.) also the ratio of the straight line B to E is given. Next, let the ratio of A to D, and of B to A B С DE F rectangle A, C to the rectangle D, F is given; and the ratio of the side A to the side D is given; therefore the ratio of the other side C to the other F is given (65. dat.). Cor. And if the extremes of four proportionals have to the extremes of four other proportionals given ratios, and one of the means a given ratio to one of the means ; the other mean shall have a given ratio to the other mean, as may be shown in the same manner as in the foregoing proposition. If four straight lines be proportionals; as the arst is to the straight line to which the second has a given ratio, so is the third to a straight line to which the fourth has a given ratio. Let A, B, C, D be four proportional straight lines, viz. as A to B, so C to D; as A is to the straight line to which B has a given ratio, so is C to a straight line to which D has a given ratio. Let E be the straight line to which B has a given ratio, and as B to E, so make D to F: the ratio of B to E.is given (Hyp.), and therefore the ratio of D to F; and because as A to B, so is C to D; and А E as B to E, so D to F; therefore, ex æquali, as A to E, so is C to F; and E is the straight line to which C D F B has a given ratio, and F that to which D has a given ratio ; therefore as A is to the straight line to which B has a given ratio, so is C to a line to which D has a given ratio. B If four straight lines be proportionals; as the first is to the straight line to which the second has a given ratio, so is a straight line to which the third has a given ratio to the fourth. * Let the straight line A be to B, as C to D: as A to the straight line to which B has a given ratio, so is a straight line to which C has a given ratio to D. Let E be the straight line to which B has a given ratio, and as B to E, so make F to C; because the А в Е ratio of B to E is given, the ratio of C to F is given : and because A is to B, as C to D; and as B F C D to E, so F to C; therefore, ex æquali in proportione perturbata (23. 5.), A is to E, as F to D; that is, A is to E, to which B has a given ratio, as F, to which C has a given ratio, is to D. * See Note, If a triangle have a given obtuse angle; the excess of the square of the side which subtends the obtuse angle, above the squares of the sides which contain it, shall have a given ratio to the triangle. Let the triangle ABC have a given obtuse angle ABC ; and produce the straight line CB, and from the point A draw AD perpendicular to BC: the excess of the square of AC above the squares of AB, BC, that is (12. 2.), the double of the rectangle contained by DB, BC, has a given ratio to the triangle ABC. Because the angle ABC is given, the angle ABD is also given ; and the angle ADB is given ; wherefore the triangle ABD is given (43. dat.) in species ; and therefore the ratio of AD to DB is given : and as AD to DB, so is (1. 6.) the rectangle AD, BC to the rectangle DB, BC; wherefore the ratio of the rectangle AD, BC to the rectangle DB, BC is given, as also the ratio of twice the rectangle DB, BC to the rectangle E H the ratio of twice the rectangle DB, BC to I G the triangle ABC is given (9. dat.); and twice the rectangle DB, BC is the excess D B с (12. 2.) of the square of AC above the squares of AB, BC; therefore this excess has a given ratio to the triangle ABC. And the ratio of this excess to the triangle ABC may be found thus : take a straight line EF given in position and magnitude ; and because the angle ABC is given, at the point F of the straight line EF, make the angle EFG equal to the angle ABC; produce GF, and draw EH perpendicular to FG ; then the ratio of the excess of the square of AC above the squares of AB, BC to the triangle ABC, is the same with the ratio of quadruple the straight line HF to HE. Because the angle ABD is equal to the angle EFH, and the angle ADB to EHF, each being a right angle; the triangle ADB is equiangular to EHF; therefore (4. 6.) as BD to DÅ, SO FH to HE ; and as quadruple of BD to DA, so is (cor. 4. 5.) quadruple of FH to HE: but as twice BD is to DA, so is (1. 6.) twice the rectangle DB, BC to the rectangle AD, BC; and as DA to the half of it, so is (cor. 5.) the rectangle AB, BC to its half the triangle ABC; therefore, ex æquali, as twice BD is to the half of DA, that is, as quadruple of BD is to DA, that is, as quadruple of FH to HE, so is twice the rectangle DB, BC to the triangle ABC. PROP. LXXV. 65. Ir a triangle have a given acute angle, the space by which the square of the side subtending the acute angle is less than the squares of the sides which contain it, shall have a given ratio to the triangle. Let the triangle ABC have a given acute angle ABC, and draw AD perpendicular to BC; the space by which the square of AC it less than the squares of AB, BC, that is, (13. 2.), the double of the rectangle contained by CB, BD, has a given ratio to the triangle ABC Because the angles ABD, ADB are each of them given, the triangle ABD is given in species; and therefore the ratio of BD to DX is given : and as BD to DA, so is A the rectangle CB, BD to the rectangle CB, AD; therefore the ratio of these rectangles is given, as also the ratio of twice the rectangle CB, BD to the rectangle CB, AD; but the rectangle CB, AD has a given ratio B D C to its half the triangle ABC ; therefore (9. dat.) the ratio of twice the rectangle CB, BD to the triangle ABC is given ; and twice the rectangle CB, BD is (13. 2.) the space by which the square of AC is less than the squares of AB, BC; therefore the ratio of this space to the triangle ABC is given: and the ratio may be found as in the preceding proposition. LEMMA. If from the vertex A of an isosceles triangle ABC, any straight line AD be drawn to the base BC, the square of the side AB is equal to the rectangle BD, DC of the segments of the base together with the square of AD; but if AD be drawn to the base produced, the square of AD is equal to the rectangle BD, DC together with the square of AB. CASE 1. Bisect the base BC in E, and join AE which will be perpendicular (8. 1.) to BC; wherefore the square of AB is equal (47. 1.) to the squares of AE, EB; but the square of EB is equal (5. 1.) to the rectangle BD, DC together with the square of DE; therefore D B DE B C A the square of AB is equal to the squares of AE, ED, that is, to (47. 1.) the square of AD, together with the rectangle BD, DC; the other case is shown in the same way by 6. 2. Elem. A If a triangle have a given angle, the excess of the square of the straight line which is equal to the two sides that contain the given angle, above the square of the third side, shall have a given ratio to the triangle. Let the triangle ABC have the given angle BAC; the excess of the square of the straight line which is equal to BA, AC together above the square of BC, shall have a given ratio to the triangle ABC. Produce BA, and take AD equal to AC; join DC, and produce it to E, and through the point B draw BE parallel to AC; join AE, and draw AF perpendicular to DC; and because AD is equal to AC, BD is equal to BE: and BC is drawn from the vertex B of the isosceles triangle DBE; therefore, by the lemma, the square of BD, that is, of BA and AC together, is equal to the rectangle DC, CE together with the square of BC; and therefore, the square of BA, AC, together, D that is, of BD, is greater than the square of BC by the rectangle DC, CE, and this rectangle has a given ratio to the F triangle ABC : because the angle BAC с is given, the adjacent angle CAD is given ; and each of the angles ADC, H DCA is given, for each of them is the G half (5.& 32.) of the given angle BAC; E therefore the triangle ADC is given K (43. dat.) in species; and AF is drawn from its vertex to the base in a given angle; wherefore the ratio of AF to the base CD is given (50. dat.); and as CD to AF, so is (1. 6.) the rectangle DC, CE to the rectangle AF, CE; and the ratio of the rectangle AF, CE to its half (41. 1.), the triangle ACE, is given ; therefore the ratio of the rectangle DC, CE to the triangle ACE, that is (37. 1.), to the triangle ABC, is given (9. dat.), and the rectangle DC, CE is the excess of the square of BA, AC together above the square of BC: therefore the ratio of this excess to the triangle ABC is given. The ratio which the rectangle DC, CE has to the triangle ABC is found thus: take the straight line GH given in posi B |