LK, KM are equal, KC is common, and KLC, KMC are right angles, CL will be equal to CM: since therefore BM is equal to BH, and CM to CL; BC will be equal to BH and CL together; and, adding AB and AC together, AB, AC, and BC will together be equal to AH and AL together : but AH, AL, are equal : wherefore each of them is equal to half the perimeter of the triangle ABC: but since AD, AE, are equal, and BD, BĚ, and also CE, CF; AB, together with FC, will be equal to half the perimeter of the triangle to which AH or AL was shewn to be equal; taking away therefore the common AB, the remainder FC will be equal to the remainder BH: in the same manner it is demonstrated, that BF is equal to CL: and since the points B, D, G, F, are in a circle, the angle DGF will be equal to the exterior and opposite angle FBH (22. 3.); wherefore their halves BĞD, HBK, will be equal to one another: the right angled triangles BGD, HBK, will therefore be equiangular, and GD will be to BD, as BH to HK: and the rectangle contained by GD, HK, will be equal to the rectangle DBH or BFC: but since AH is to HK, as AD to DG, the rectangle HAD (22. 6.) will be to the rectangle contained by HK, DG, or the rectangle BFC (as the square of AD is to the square of DG, that is), as the square of the radius is to the square of the tangent of the angle DAG, that is, the half of BAC: but HA is half the perimeter of the triangle ABC, and AD is the excess of the same above HD, that is, above the base BC; but BF or CL is the excess of HA or AL above the side AC, and FC, or HB, is the excess of the same HA above the side AB; therefore the rectangle contained by half the perimeter, and the excess of the same above the base, viz. the rectangle HAD, is to the rectangle contained by the straight lines by which the half of the perimeter exceeds the other two sides, that is, the rectangle BFC, square of the radius is to the square of the tangent of half the angle BAC opposite to the base. as the Q. E. D. PROP. VII. Fig. 12. 13. In a plane triangle, the base is to the sum of the sides as the difference of the sides is to the sum or difference of the segments of the base made by the perpendicular upon it from the vertex, according as the square of the greater side is greater or less than the sum of the squares of the lesser side and the base. Let ABC be a plane triangle; if from A the vertex be drawn a straight line AĎ perpendicular upon the base BC, the base BC will be to the sum of the sides BA, AC, as the difference of the same sides is to the sum or difference of the segments CD, BD, according as the square of AC the greater side is greater or less than the sum of the squares of the lesser side AB, and the base BC. About A, as a centre, with AC the greater side for a distance, let a circle be described meeting AB produced in E, F, and CB in G: it is manifest, that FB is the sum, and BE the difference of the sides; and since AD is perpendicular to GC, GD, CD, will be equal; consequently GB will be equal to the sum or difference, of the segments CD, BD, according as the perpendicular AD meets the base produced, or the base; that is (by Conv. 12. 13. 2.) according as the square of AC is greater or less than the sum of the squares of AB, BC: but (by 35. 3.) the rectangle CBG is equal to the rectangle EBF; that is (16. 6.) BC is to BF, as BE is to BG; that is, the base is to the sum of the sides, as the difference of the sides is to the sum or difference of the segments of the base made by the perpendicular from the vertex, according as the square of the greater side is greater or less than the sum of the squares of the lesser side and the base. Q. E. D. PROP. VIII. PROB. Fig. 14. The sum and difference of two magnitudes being given, to find them. Half the given sum added to half the given difference, will be the greater, and half the difference subtracted from half the sum, will be the less. For let AB be the given sum, AC the greater, and BC the less. Let AD be half the given sum; and to AD, DB, which are equal, let DC be added; then AC will be equal to BD and DC together: that is, to BC, and twice DC; consequently, twice DC is the difference, and DC half that difference; but AC the greater is equal to AD, DC; that is, to half the sum added to half the difference, and BC the less is equal to the excess of BD, half the sunı, above DC half the difference. R. E. D. SCHOLIUM. Of the six parts of a plane triangle (the three sides and three angles) any three being given, to find the other three is the business of plane trigonometry; and the several cases of that problem may be resolved by means of the preceding propositions, as in the two following with the tables annexed. In these, the solution is expressed by a fourth proportional to three given lines; but if the given parts be expressed by numbers from trigonometrical tables, it may be obtained arithmetically by the common Rule of Three. Note. In the tables the following abbreviations are used: R. Is put for the Radius ; T. for Tangent; and S. for Sine. Degrees, minutes, seconds, &c. are written in this manner; 30° 25' 13', &c. which signifies 30 degrees, 25 minutes, 13 seconds, &c. Solution of the Cases of Right Angled Triangles. GENERAL PROPOSITION. angles, any two being given besides the right angle, the It is manifest from 47. 1. that of the two sides and hypothenuse, if any two be given, the third may also be found. It is also manifest from 32. 1. that if one of the acute angles of a right angled triangle be given, the other is also given, for it is the complement of the former to a right angle. If two angles of any triangle be given, the third is also given, being the supplement of the two given angles to two right angles. The other cases may be resolved by help of the pre- Fig. 15. ceding propositions, as in the following table: Given. Sought. 1 Twosides, AB, The angles AB: AC:: R:T, B, of AC. B, C. which Cis the complement. 2 AB, BC, aside The angles BC:BA::R:S, C, of and the hypo-B, C. which Bisthe complement. thenuse. 3 AB, B, a side The other R:T, B:: BA: AC. and an angle. side AC. 4 AB and B, a The hypo- S, C:R:: BA : BC. 5 BC and B, the The side R:S, B:: BC: CA. These five cases are resolved by Prop. 1. H H Solution of the Cases of Oblique Angled Triangles. GENERAL PROPOSITION. In an oblique angled triangle, of the three sides and three angles, any three being given, the other three may be found, except when the three angles are given; in which case the ratios of the sides are only given, being the same with the ratios of the sines of the angles opposite to them. Fig. 16. 17. Given. Sought. 1 A,B,and there- BC, AC. S, C:S, A:: AB: BC, and also S,C: S,B:: AB : AC. (2.) 2 AB, AC, and The angles AC: AB::S,B: S,C. B, two sides A and C. (2.) This case admits oftwo and an angle solutions; for C may be opposite to one greater or less than a quaof them. drant. (Cor. to def. 4.) 2 2 ed angle. 3 AB, AC, and The angles AB+AC: AB-AC::T, : T, (3.) the sum and difference of the angles C, B, being given, each of them is given. (7.) Otherwise. Fig. 18. BA: AC::R:T,ABD, and also R:T,ABD-45° B+C B-C ::T, (4.) therefore B and Care given as before. (7.) :T, 2 2 |