If there be another proportion G: H:: E: F, then it follows, from what has been proved, considering A+ C, as a single magnitude, and likewise B + D, that A+C+G: B + D + H :: E: F; and so on, whatever be the number of proportions having the same two last terms in each. COR. Hence, since two similar parallelograms are composed of two pairs of similar triangles, which are to each other as the squares of corresponding sides or lines within them (76), therefore the parallelograms also are proportional to the squares of their homologous sides or other corresponding lines within them. 81. PROP. XIV. If any two chords be drawn in the same circle intersecting each other, the rectangles contained by the parts into which each is divided by the point of intersection are equal to one another. Let AB, CD be two chords of a given circle intersecting in E. Then the rectangle AE, EB shall be equal to the rectangle CE, ED. Join AC, BD. Then ACD A = LABD, being angles in the same segment (52 Cor.), that is, LACE=LEBD. Similarly, CAE = LBDE. Also AEC-¿BED (31); .. the triangles AEC, BED, L L B are similar; and .. the sides about equal angles are proportionals; that is, AE: EC:: ED : EB (71), and.. the rectangle AE, EB = the rectangle CE, ED (74). COR. 1. If one of the chords, AB, bisect the other CD, in the point E, then it follows that the rectangle AE, EB is equal to the square of CE. COR. 2. If AB bisect CD at right angles, AB will be a diameter, and then also the rectangle AE, EB = the of CE. square 82. PROP. XV. If any two chords of the same circle be produced to meet in a point without the circle, the rectangles contained by the whole line and the part produced, for each chord, shall be equal to one another. Let AB, CD be two chords of the same circle, which produced meet in the point P; the rectangle PA, PB shall be equal to the rectangle PC, PD. Join AC, BD; then since ABDC is a quadrilateral inscribed' in a circle, ABD + L ACD = two right angles (53), = ¿ACP + ¿ACD (30), .. ▲ABD = LACP. And angle at Pis common to the two triangles PBD, PAC; .. remaining angle PAC = remaining angle PDB; and.. the triangles PBD, PAC are similar, and the sides about D equal angles proportionals (71), .. PA: PC:: PD: PB, and.. the rectangle PA, PB = the rectangle PC, PD (74). COR. If one of the chords, AB, be a diameter, the proposition holds true, viz. that the rectangle PA, PB =the rectangle PC, PD. 83. PROP. XVI. If any chord and tangent of the same circle be produced until they meet, the rectangle contained by the whole chord thus produced and the part produced shall be equal to the square of the tangent, that is, of the line between its point of intersection with the chord produced and the point where it touches the circle. Let AB be a chord, and PC a tangent at the point C, of the circle ABC, whose centre is 0; and let BA produced meet PC in the point P. Then the rectangle PA, PB shall be equal to the square of PC. Join PO, and produce it to the circumference of the circle, so that it meets the circumference in E and F, making EF a diameter. With centre O and radius OP de B scribe another circle, that is, concentric with the former; join OC, and produce it both ways to meet the outer circumference in G and H, so that GH is a diameter; and produce PC to meet this circumference in D, so that PD is a chord of the outer circle. Then, since GH is a diameter at right angles (55 Cor. 1) to the chord PD, PD is bisected in C (49 Cor.); and .. (81 Cor. 2) the rectangle GC, CH= the square of PC. But GC PF, and CH PE; .. the rectangle PE, PF the square of PC. But the rectangle PE, PF = the rectangle PA, PB (82), = = = ..the rectangle PA, PB = the square of PC. 84. PROP. XVII. In the same circle, or in equal circles, any two arcs are proportional to the angles which they subtend at the centre. Then Let AB, CD be any arcs of two equal circles; find the centres E, F, and join EA, EB, FC, FD. arc AB arc CD :: 4 AEB: 4 CFD. For, assuming that there is some small arc which, taken as the unit of measurement, is contained an exact number of times both in AB and CD, let Aa be such arc, and suppose it to be contained 5 times in AB, and 3 times in CD, (the proof is the same whatever the numbers be); draw the radii, as Ea, and Fc, to the several points of division in AB and CD, so that the arc AB is divided into 5 equal parts, CD into 3 equal parts, and also the angles AED, CFD, into 5 and 3 equal angles, respectively, since equal arcs in the same circle, or in equal circles, subtend equal angles at the centre (59). Then, since arc AB contains a certain unit 5 times, and arc CD the same unit 3 times, arc AB : arc CD :: 5 : 3. Again ▲ AED contains 4 AEa 5 times, and CFD contains CFC, which is equal to AEa, 3 times, .. LAEB: 4 CFD :: 5 : 3. And. arc AB: arc CD :: 4 AEB: 4 CFD. NOTE. It will have been noticed, that, in the preceding Theory of Ratio and Proportion, the magnitudes compared are assumed to be, what is called, 'commensurable', that is, to have a 'common measure', or common unit of measurement. Now two or more magnitudes are said to have a 'common measure', when each of them contains the unit of measurement a certain number of times exactly without remainder. Thus two lines, which are 5 yards and 73 yards in length respectively, are commensurable, because, taking the foot as the measure, the first line contains it 16 times, and the second 23 times, exactly. Similarly, two lines which are 2 yards and 1 yards in length respectively, are commensurable, because the first contains an inch 90 times, and the second 54 times, exactly. But it does not follow, (and in fact it is not true,) that all lines are of this kind, that is, commensurable. Lines, and also areas, have sometimes to be compared, which have no common measure, and are called incommensurable. To these the preceding Theory does not with perfect mathematical accuracy apply, as it does to commensurable magnitudes; although in all such cases a measure may be found which shall approach as nearly as we please to a common measure, and thus render the preceding Theory applicable by approximation, and to all practical purposes sufficiently true. Euclid's method of treating ratios and proportion, which applies strictly and equally to all magnitudes, commensurable and incommensurable, has not been adopted, simply because it does not admit of being presented in a form sufficiently intelligible to those for whom this little work is designed. It seemed better to employ a method, which, with admitted imperfections, would allure the learner, than to aim at a perfectness of theory, which might lead him either to pass over the subject entirely, or to read it and not understand it. EXERCISES C. (1) Define 'ratio'; between what sort of magnitudes can it exist? Is there any ratio' between ten shillings and two miles? If not, why not? (2) What is the test by which you determine whether, or not, two proposed magnitudes are of the same kind'? Apply it to the case of a triangle and one of its sides. Also to the case of a triangle and a square, (3) Is there any ratio between an angle and a triangle? Or between a right angle and a square ? (4) Define the measure of a ratio; and express the ratio of a parallelogram to the triangle' on the same base and between the same parallels'. (5) What is the ratio of a lineal inch to a lineal yard? (6) What is the ratio of the square of AB to the square of the half of AB? (7) Is it necessary, when two lines or magnitudes have a ratio to each other, that the one should contain the other an exact integral number of times? Explain fully. (8) When is one line, or area, said to be a multiple of another? If 5 times A = 7 times B, what is the ratio of A to B? (9) Define 'proportion', and 'proportional'. How many magnitudes are concerned in a proportion? May they be all of one kind? Must they be so? (10) Can two lines and two triangles be in proportion? Can two angles, a triangle, and a parallelogram, be in proportion? (11) If there be two triangles of equal altitudes, and the base of one be double the base of the other, what is the proportion between the bases and triangles? And what is the ratio of the two triangles? (12) Shew that, if any two sides of a triangle be bisected, the line joining the points of bisection is parallel to the third side, and equal to half of it. |