A Let A, and D, be so placed that two sides of one square may be in the same straight lines with two sides of the other, each to each; and produce the other sides of A and D until they meet. The resulting fig, will be a square, composed of the two squares A and D, together with two equal rectangles, E, E. Then since A and E are parallelograms of the same altitude, base of A: base of E. A: E Also E: D:: base of A: base of E; .. A: E :: E : D. The same construction being made with B and C, as in the annexed fig., it may be shewn in the same manner that B: F :: F: C. Now let a, b, c, d, e, f represent straight lines having the same ratio to each other as the areas A, B, C, D, E, F. Then, since A: B:: a: b, CD cd, and A: B :: C : D, E d = E D .. a b c d; and therefore the rectangle contained by a and d = the rectangle contained by b and c. Similarly, aeed, and b:ƒ::ƒ: c; .. rectangle a, square of e, and rectangle b, c= square of f; .. square of e = square off, and .. e=ƒ (42 Cor. 3). But E: Fe: f, .. E = F. Now E is the rectangle contained by a side of A and a side of D; and F is the rectangle contained by a side of B and a side of C; and rectangle E = rectangle F; therefore (74 Cor. 1), side of A side of B :: side of C: side of D. COR. 1. Hence also the converse is easily shewn, viz. that, if four straight lines are proportionals, the squares described upon them are proportionals. COR. 2. In (76) it was proved that the areas of similar triangles are to one another as the squares of corre may be done here without inconvenience, because we are not much concerned with the magnitude of the sides until we arrive at the last stage of the proof. sponding, or homologous', sides. It may now be shewn that the areas are proportional to the squares of the altitudes also of the triangles, or to the squares of any corresponding lines within the triangles. For (see fig. in 76) CD: cd BC: bc = AB: ab, square of CD: square of cd = square of AB : square of ab; and .. triangle ABC : triangle abc square of CD: square cd. = 80. PROP. XIII. If there be any number of magnitudes, which, taken two and two, have a certain fixed ratio to each other, the sum of the first terms of the several pairs of magnitudes shall be to the sum of their second terms in that same ratio. Let A, B, C, D be four magnitudes such that A: B :: E: F, and C: D:: E: F, then also A+C: B+ D :: E : F. For suppose a, b, c, d, e, f to represent six straight lines having the same ratio to each other that A, B, C, D, E, F, have. Then since a:be:f, the rectangle a, f= the rectangle b, e (74). Similarly, the rectangle c, f= the rectangle d, e. .. rectangle a, ƒ + rectangle c, f= rectangle b, e + rectangle d, e. H Now, constructing each of these rectangles, as in the annexed figs. by making AB= a, BC=c, CD=f; also EF = b, FG=d, GH =e; it is obvious that the rectangle a, ƒ+ rectangle c, f= rectangle contained by AB + BC, and CD, that is, rectangle a +c, f. B E F G Also rectangle b, e + rectangle d, e = rectangle contained by EF + FG, and GH, that is, rectangle b+d, e. .. rectangle a + c, f= rectangle b + d, e, But a+c: b+ d :: A + B : B + D, and ef E: F, by supposition, ..A+C: B+D: E: F. 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. |