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24 Th. If the first be to the second as the third is to the fourth, and the fifth to the second as the sixth to the fourth ; then the sum of the first and fifth is to the second, as the sum of the third and sixth is to the fourth.
Take six magnitudes AB, C, DE, F, BG, EH, in order. Then, by hyp. AB: C:: DE: F
and BG:C:: EH:F; and it is to be proved that AB+BG:C:: DE+EH:F.
Now, since BG:C::EH:F
AB: C:: DE: F
AB : BG:: DE: EH (). Again,
AB+BG:C:: DE+EH: F (6).
AB-BG:C:: DE-EH: F. P. 19 of b 5. Cor. 2. And in any two ranks of magnitudes, how many soever they may be in each rank, if each term of the first have to a magnitude the same ratio as each term of the second has to another magni. tude, the sum of the first rank has to its magnitude the same ratio which the sum of the second rank has to its magnitude.
25 Th. If four magnitudes of the same kind be proportionals, the sum of the greatest and least of them is greater than the sum of the other two.
Let AB, CD, E, F be four proportionals, in order, namely, AB : CD::E: F. And because they are of the same kind (a), they are measured by the same unit, and are terms of the same numerical series: therefore, the extremes in position are the extremes in magnitude.
Let AB be the greatest and F the least. In AB take AG=E, and in CĎ take CH=F. Then,
AB: CD :: AG: CH. The remaining parts, GB, HD, have also the same ratio as the wholes (b); therefore AB : CD :: GB : HD. But AB is greater than CD; therefore GB is greater than HD (c).
And because AG=E, and CH=F; therefore AG+F=CH+E (d). To the left of this equation add GB, the greater; to the right add HD, the less: therefore AG+GB+F exceeds CH+HD+E (e); that is, AB+F is greater than CD+E. Wherefore, if four magnitudes, &c.
Q. E. D. Recite (a) def. 4, 5; (6) p. 19, 5; (c) p. A, 5;
(d) ax. 2, 1; (e) ax. 4, 1.
F Th. Ratios composed of equal ratios are equal to each other.
Let the ratios of A to B and of B to C be severally equal to the ratios of D to E, and of E to F; the ratio of A to C equals that of D to F.
For, let A to B equal D to E, and B to C equal E to F; then A is to coas D is to F-ex æquo (a). Or, let A to B equal E to F,
B to C equal D to E; then A is to C as D is to F-ex æquo perturbato (6).
Wherefore, ratios composed (or compounded), &c. Q. E. D. Recite (a) p. 22, 5; (6) p. 23, 5.
G Th. If the same unit measure two magnitudes, it will also measure their sum; and if the magnitudes be unequal, it will measure their difference.
For since the same unit measures the two magnitudes, they are of the same kind, and the sum of their units may be placed in a consecutive series, to the whole of which the same measure applies. Again, if one be greater than the other, the difference is a multiple of the same measuring unit. Therefore, if the same unit, &c.
Q. E. D.
1. Similar rectilineal figures are those which have their angles equal, each to each; and the sides about the equal angles proportionals.
2. Reciprocal figures, viz. triangles and parallelograms, are those which have their sides about two of their angles proportionals, in such manner, that the extremes belong to one of the figures, and the means to the other; and such figures are equal to one another.
3 A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment, as the greater segment to the less.
4. The altitude of any figure is the distance between two parallel lines touching the figure, one of which is its base.
1 Th. Triangles and parallelograms of the same, or equal altitudes, are to each other as their bases
Given the triangles ABC, ACD, and the parallelograms CE, CF, on the bases BC, CD, and be. tween two parallels (a); then ABC is to ACD, and CE is to CF, as BC is to CD.
Take HC, CL, equimultiples of BC, CD; and divide CH, CL into parls equal to CB, CD: then, because each of the parts GB, GH equals BC; and each of the parts KD, KL equals CD; therefore CH and CL are 3d multiples of BC and CD (6). To the point A join the points H, G, K, L: then the triangles ABC, AGB, AHG are all equal; and likwise the triangles ACD, ADK, AKL (c); and the triangles ACH, ACL are 3d multiples of the triangles ABC, ACD: and if the base CH be greater than the base CL, the triangle ACH is greater than the triangle ACL; if equal, equal ; and if less, less. (d).
Now the bases BC, CD, and the triangles ABC, ACD, are four magnitudes; and of the 1st BC, and 3d ABC, equimultiples CH, ACH, are taken; and of the 2d and 4th CD, ACD, equimultiples CL, ACL, are taken; also it is shown, that if the base CẢ be greater than the base CL, the triangle ACH is greater than the triangle ACL; if equal,
equal; and if less, less: therefore, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD (d).
But the parallelograms CE, BF are 2d multiples of the triangles ABC, ACD (e), and the equimultiples have the ratio of their magnitưdes (f); therefore, ABC : ACD :: CE: CF. And, since ABC: · ACD :: BC: CD; therefore (8) also CE: CF :: BC : CD. Wherefore, triangles, &c.
Q. E. D. Recite (a) def. 4, 6; (b) def. A, 5; (c) p. 38, 1;
(d) def. 5,5; (e) p. 41, 1; (f) p. 15, 5;
(8) p. 11, 5. Cor. It may be also proved (in the same way,) that triangles and parallelograms of the same or equal bases are to each other as their altitudes.
2 Th. If a straight line be drawn parallel to one side of a triangle, it shall cut the other sides, or those sides .produced proportionally: and if the sides, or the sides produced, be cut proportionally, the straight line joining the sectional points, shall be parallel to the third side of the triangle.
Given the triangle ABC, and DE drawn parallel to BC, cutting AB, AC; 1, within the triangle; 2, below the base; 3, above the vertex : in either case, BD: AD: : CE: AE. Join BE, CD.
1. The triangles BDE, CDE are equal; D being on the same base DE, and between the same parallels BC, DE (a): but ADE is another triangle, to which the equals B BDE, CDE havethe same ratio (6); and BDE: ADE:: BD: AD (c); also, CDE : ADE:: CE : AE (c): therefore BD is to AD as CE is to AE (d).
2. Upon the same construction. If BD : AD :: CE: AE, then DE is parallel to BC. For BD: AD:: BDE : ADE; and CE: AE :: CDE : ADE (c): therefore, BDE and CDE having the same ratio to ADE are equa' (b); and being on the same base DE, and same side of it, they are also between the same parallels (e): therefore DE is parallel to BC. Wherefore, if a straight line, &c.
Q. E. D. Recite (a) p. 37,
1; (b) p. 7, 9, of b. 5; (c) p. 1, 6; (d) p. 11, 5; (e) p. 39, 1.
3 Th. If a straight line bisect an angle of a triangle, and also cut the base ; the segments of the base shall have the same ratio as the sides which contain the angle: and
if the segments of the base have the same ratio as the other sides, the straight line drawn fronı the vertex shall bisect the vertical angle.
Given the triangle ABC, and AD bisecting the angle A : then BD: DC :: BA : AC;--and, So, the angle A is bisected by AD.
Through C draw CE parallel to AD (a); and produce BA to meet CE in E.
1. Because BE and AC meet the parallels AD, CE, the exterior angle BAD, and the interior BEC, or AEC are equal(6), also the alternate angles DAC, ACE (c): but BAD equals DAC, by hypothesis; therefore AEC eqnals ACE (). and the side AE equals the side AC (c). And because AD is drawn parallel to CE; therefore, BD: DC :: BA: AE, ($ ), or its equal AC.
2. Because BD: DC :: BA : AC, by hyp. and, that from the parallels, BD: DC::BA: AE (S); therefore AE equals AC (8); and their opposite angles AEC, ACE are equal (h): but AEC als BAD (), and DAC equals ACE; therefore BAD equals DAC (i), and so, the angle BAC is bisected by AD. Wherefore, it a straight line bisect, &c.
Q. E. D. Recite (a) p. 31, 1;
(b) p. 29, 1; (c) p. 27, 1; (d) ax. 1, 1; (e) p. 6, 1; (f) p. 2,6; (g) p. 9, 5; (h) p. 5, 1; (i) ax. 1, 1.
A Th. If a straight line bisect the exterior angle of a triangle and meet the base produced, the segments between the meeting point and each extremity of the base shall have the same ratio as the other sides of the triangle. And if the segments of the base produced have the same ratio as the other sides, the straight line drawn from the vertex bisects the exterior angle.
Given the triangle ABC, its side BA produced to E, the exterior angle CAE bisected
E by AD which meets BC produced in D. BD: DC :: BA: AC; and if so, the exterior angle CAE is bisected by AD.
F Through C draw CF parallel to AD (a).
1. Because the straight lines AC, FE meet the parallels AD, CF, the alternate angles DAC, B ACF are equal (b), and so are the interior and exterior AFC, DAE (c): but by hyp. DAC equals DAE; therefore ACF equals AFC (d), and AF equals AC (e): and because CF is drawn parallel to AD; therefore BD: CD:: BA : AF ($), or AC its equal (f).