ones, (for EDB is = FBD (= right 'angle) because DE is parallel to FB) it is plain, that, if BD be made the radius, BF will be the tangent of BDF, and DE the tangent of DBE but, because of the similar triangles CFB and CDE, CF: CD:: BF: DE; that is, as the sum of the sides AC and AB, is to their difference: so is the tangent of BDF, to the tangent of DBC; which angle is, manifestly, the excess of ABC, above BDF, or ABD; and also the excess of ADB above ACB. Q. E. D. PROPOSITION XI. As the base of any plane triangle is to the sum of the two sides, so is the difference of the sides to the difference of the segments of the base, made by a perpendicular falling from the vertical angle. For, Let ABC be the proposed triangle, and BD the perpendicular; from B as a centre, with the interval BC, let the circumference of a circle be described, cutting the base AC in G and the side AB, produced, in F and E: then will AE be the sum of the sides, AF their difference, and AG the difference of the segments of the base AD and DC: but (by Euc. 36. 3.) AEX AF= AEXAF= A AC x AG; and therefore AC: AE:: AF: AG. Q. E. D. E The Solution of the cases of right-angled plane triangles. A C JB Case Given. The hypothe Sought. 1 nuse AC and The leg BC. the angles. Proportion. As the radius (or the sine of B) is to the hyp. AC; so is the sine of A, to its opposite side BC. (by prop. 9.) As AC rad. :: AB : sine of C; whose complement gives the angle A. Let the angles be found by case 2; then, as rad. : AC:: sine of A: BC. (by prop. 9.) As sine of C: AB:: rad. (sine of B): AC. (by prop.) As sine of C: AB:: sine of A: BC (by prop. 9.) Or, rad. : tang. of A:: AB: BC. (by prop. 8.) As AB: BC: : rad. : tang. off A (by prop. 8); whose complement gives the angle C. Find the angles, by case 6, and from thence the hyp. AC, by case 4. SECTION XVIII. The Application of Algebra to the Solution of Geometrical Problems. WHEN a geometrical problem is proposed to be resolved by algebra, you are, in the first place, to describe a figure that shall represent, or exhibit the several parts or conditions thereof, and look upon that figure as the true one; then, having considered attentively the nature of the problem, you are next to prepare the figure for a solution (if need be) by producing, and drawing, such lines therein as appear most conducive to that end. This done, let the unknown line, or lines which you think will be the easiest found (whether required or not,) together with the known ones (or as many of them as are requisite,) be denoted by proper symbols; then proceed to the operation, by observ ing the relation that the several parts of the figure have to each other; in order to which, a competent knowledge in the elements of geometry is absolutely necessary. As no general rule can be given for the drawing of lines, and electing the most proper quantities to substitute for, so as to always bring out the most simple conclusions (because different problems require different methods of solution,) the best way, therefore, to gain experience in this matter is to attempt the solution of the same problem several ways, and then apply that which succeeds best to other cases of the same kind, when they afterwards occur. I shall, however, subjoin a few general directions which will be found of use. 1o. In preparing the figure, by drawing lines, let them be either parallel or perpendicular to other lines in the figure, or so as to form similar triangles; and if an angle be given, let the perpendicular be opposite to that angle, and also fall from the end of a given line, if possible. 2o. In electing proper quantities to substitute for, let those be chosen (whether required or not) which lie nearest the known or given parts of the figure, and by help whereof the next adjacent parts may be expressed, without the intervention of surds, by addition and subtraction only. Thus, if the problem were to find the perpendicular of a plane triangle, from the three sides given, it will be much better to substitute for one of the segments of the base, than for the perpendicular, though the quantity required; because the whole base being given, the other segment will be given, or expressed, by subtraction only, and so the final equation come out a simple one; from whence the segments being known, the perpendicular is easily found by common arithmetic: whereas, if the perpendicular were to be first sought, both the segments would be surd quantities, and the final equation an ugly quadratic one. 3o. When, in any problem, there are two lines or quantities alike related to other parts of the figure, or problem, the best way is to make use of neither of them, but to substitute for their sum, their rectangle, or the sum of their alternate quotients, or for some line or lines in the figure, to which they have both the same relation. This rule is exemplified in prob. 22, 23, 24, and 27. 4°. If the area, or the perimeter of a figure be given, or such parts thereof as have but a remote relation to the parts required, it will sometimes be of use to assume another figure similar to the proposed one, whereof one side is unity, or some other known quantity; from whence the other parts of this figure, by the known proportions of the homologous sides, or parts, may be found, and an equation obtained, as is exemplified in prob. 25 and 32. These are the most general observations I have been able to collect; which I shall now proceed to illustrate by proper examples. PROBLEM I. The base (b), and the sum of the hypothenuse and perpendicular (a) of a right-angled triangle ABC, being given; to find the perpendicular. |