0 g. da': to L are given, the ratio of H to L is given b; because, there. fore, as AEB to FD, so is H to K; and as FD to AG, so is K to L; ex aequali, as AEB to AG, so is H to L; there. fore 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 fimilar to AG: Let this be the parallelogram MN; MN also is fimilar to FD; for, by the con. struction, MN is similar to AG, and AG is similar to FD; therefore the parallelogram FD is similar to MN. IF given ratios to the extremes of other three propor, tional straight lines; the means shall also have a given ratio to one another : And if one extreme has 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 a 6y. dat. ratio of the rectangle A, C to the rectangle D, F is given; b 17.6. but the square of B is equal to the rectangle A, C; and the square of E to the rectangle b D, F; therefore the ratio of the c58. dat. square of B to the square of E is given ; wherefore also the ra tio of the straight line B to E is given. Next, let the ratio of A to D, and of B to E be gi- АВС d 54. dat. the square of B to the square of E is givend; therifore the ratio of the rectangle A, C to the rectangle DEF D, F is given ; and the ratio of the side A to the lide D is given ; therefore the ratio of the other fide C to € 65. dat. the other F is given. 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 mçan, as may be shown in the fame manner as in the foregoing proposition. PROP. IF 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 itraight 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', and therefore the ratio of D to F; a Hyp. and because as A to B, so is C to D; and as B to E fo D to F; therefore, ex aequali, as A to E, fo is A BE C to F; and E is the straight line to which B has a CDF 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. IF the straight line to which the second has a given ratio, so is a Atraight 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 Atraight line to which B has a given ratio, fo is a straight line to which C has a giver ratio to D. Let E be the itraight line to which B has a given ratio, and as B to E, so make F to C; because the 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 A B E to E, fo F to C; therefore, ex aequali in proportione F CD perturbata', 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 has a given ratio, is to D. 13. Si IF. a triangle has a given obtufe 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. 2 12. 2. 9. dat, Let the triangle ABC have a given obtuse angle ABC; and produce the straight line CB, and from the point A dras AD perpendicular to BC: The excess of the square of AC a. bove the squares of AB, BC, that is “, 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 gi ven; and the angle ADB is given ; wherefore the triangle 6 43. dat. ABD is given in species; and therefore the ratio of AD to C1.6. DB is given : And as AD to DB, so is the rectangle AD, E HI the ratio of twice the rectangle DB, BC FG to the triangle ABC is given; and twice the rectangle DB, BC is the excess a of D B the square of AC above the squares of AB, Bư; 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 Itraight 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 trif 4. 6. angle ADB is equiangular to EHF; thereforef as BD to DA, 8 Cor. 4. 5. fo FH to HE; and as quadruple of BD to DA, so is 8 qua druple of FH to HE: But as twice BD is to DA, so is twice the rectangle DB, BC to the rectangle AD, BC; and as DA C. 5 to the half of it, so is the rectangle AD, BC to its half the triangle triangle ABC; therefore, ex aequali, as twice BD is to the half of DA, that is, as quadruple of BD is to DA, that is, as qua. druple of FH to HE, fo is twice the rectangle DB, BC to the triangle ABC. IF a triangle has 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 is less than the squares of AB, BC, that is, the double a 13. 2, 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 fpecies ; and therefore the ratio of BD to DA is given : And as BD to DA, А so is the rectangle CB, BD to the rectangle CB, AD; therefore the ratio of these rečt. angles 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 to its balf the triangle ABC, therefore the B D C 0 9. dako satio of twice the rectangle CB, BD to the triangle ABC is given; and twice the rectangle CB, BD is 'the space by which the square of AC is less than the squares of AB, BC, there. fore the ratio of this space to the triangle ABC is given : And the ratio may be found as in the preceding proposition. L E M M A. IF from the vertex A of an isosceles triangle ABC, any straight line AD be drawn to the base BC, the square of the Gide 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 iquare of AD is equal to the rectangle BD, DC together with the square of AB. CAS. 1. Bifeet the bare BC in E, and A join AE which will be perpendicular a to a 8. s. BC; wherefore the square of AB is equal to the squares of AE, EB ; but the square b 4. 7. I of EB is equal to the rectangle BD, DC C 5. I. together with the square of DE; therefore the square of AB is equal to the D BDE C Г.ua: $ 47. I. squares of AE, ED, that is, to the square of AD, together with the rectangle BD, DC; the other case is shown in the fame way by 6. 2. Elem. 67. PROP. LXXVI. Fa 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 fide, 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 be. Cause AD is equal to AC, BD is equal to BE; and BC is drawn from the vertex B of the isosceles triangle DBE, cherefore, 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, that is, of BD, is greater than the square of BC by the rectangle DC, D CE; and this rectangle has a given F ratio to the triangle ABC: Because the angle BAC is given, the adjacent B с angle CAD is given; and each of the angles ADC, DCA is given, for each 25.& 32. I. of them is the half of the given angle G E BAC; therefore the triangle ADC is b 43. dat. given in species; and AF is drawn from its vertex to the base in a given angle; wherefore the ratio c 50. dat. of AF to the base CD is given; and as CD to AF, so is the d1. 6. rectangle DC, CE to the rectangle AF, CE; and the ratio of the rectangle AF, CE to its half the triangle ACE is given ; therefore the ratio of the rectangle DC, CE to the triangle f 37. I. ACE, that is f, to the triangle ABC, is given &; and the rectangle & 9. dat. 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 poli C 41. I. |