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lipder GQ; and it has been demonstrated, if the axis KL Book XII. be greater than the axis KM, the cylinder PG is greater than the cylinder GQ; and if equal, egual; and if less, less : Therefored the axis EK is to the axis KF, as the cylinder « 5 Def. 5. BG to the cylinder GD. Wherefore, if a cylinder, &c. Q.E.D.
PROP. XIV. THEOR.
Cones and cylinders upon equal bases are to one another as their altitudes.
Let the cylinders EB, FD, be upon equal bases AB, CD: As the cylinder EB to the cylinder FD, so is the axis GH to the axis KL.
Produce the axis KL to the point N, and make LN equal to the axis GH, and let CM be a cylinder of which the base is CD, and axis LN; and because the cylinders EB, CM, have the same altitude, they are to one another as their bases 8: But their bases are equal, therefore also the cylin- . 11. 12. ders EB, CM, are equal : And because the cylinder
F FM is cut by the plane CD parallel to its opposite planes, as the cylinder CM to the E
G cylinder FD, so is b the axis
b 13. 12. LN to the axis KL: But the cylinder CM is equal to the cylinder EB, and the axis LN to the axis GH: Therefore as the cylinder EB to the cylin- 1 H
M der FD, so is the axis GH to the axis KL: And as the cylinder EB to the cylinder FD, so is the cone ABG to the cone CDK, because the cylin- c 15. 5. ders are tripled of the cones : Therefore also the axis GH a 10. 19. is to the axis KL, as the cone ABG to the cone CDK, and the cylinder EB to the cylinder FD. Wherefore cones, &c. Q.E.D.
PROP. XV. THEOR.
See N. The bases and altitudes of equal cones and cylin
ders are reciprocally proportional; and if the bases
Let the circles ABCD, EFGH, the diameters of which
Either the altitude MN is equal to the altitude KL, or these altitudes are not equal. First, let them be equal; and the cylinders AX, EO being also equal, and cones and cy
linders of the same altitude being to one another as their * 11. 12. bases a, therefore the base ABCD is equalb to the base
EFGH; and as the base ABCD is to the base EFGH, so
YK | P1 Xs
MP: And because the cylinder AX is equal to the cylinder €7. 5. EO, as AX is to the cylinder ES, so is the cylinder EO to
the same ES: But as the cylinder AX to the cylinder ES,
ders AX, ES are of the same altitude ; and as the cylinder • 13. 12. EO to the cylinder ES, sod is the altitude MN to the alti
tude MP, because the cylinder EO is cut by the plane TYS
parallel to its opposite planes. Therefore as the base ABCD Book XII. to the base EFGH, so is the altitude MN to the altitude MP: But MP is equal to the altitude KL; wherefore as the base ABCD to the base EFGH, so is the altitude MN to the altitude KL; that is, the bases and altitudes of the equal cylinders AX, EO, are reciprocally proportional.
But let the bases and altitudes of the cylinders AX, EO be reciprocally proportional, viz. the base ABCD to the base EFGH, as the altitude MN to the altitude KL: The cylinder AX is equal to the cylinder EO.
First, Let the base ABCD be equal to the base EFGH; then because as the base ABCD is to the base EFGH, so is the altitude MN to the altitude KL; MN is equal b to KL, A. 5. and therefore the cylinder AX is equal a to the cylinder EO.. 11.12.
But let the bases ABCD, EFGH be unequal, and let ABCD be the greater; and because as ABCD is to the base EFGH, so is the altitude MN to the altitude KL; therefore MN is greater b than KL. Then, the same construction being made as before, because as the base ABCD to the base EFGH, so is the altitude MN to the altitude KL; and because the altitude KL is equal to the altitude MP; therefore the base ABCD is a to the base EFGH as the cylinder AX to the cylinder ES; and as the altitude MN to the altitude MP or KL, so is the cylinder EO to the cylinder ES: Therefore the cylinder AX is to the cylinder ES, as the cylinder EO is to the same ES: Whence the cylinder AX is equal to the cylinder EO; and the same reasoning holds in cones. Q.E. D.
PROP. XVI. PROB.
To describe in the greater of two circles that have the same centre, a polygon of an even number of equal sides, that shall not meet the lesser cirele.
Let ABCD, EFGH be two given circles having the same centre K: It is required to inscribe in the greater circle ABCD, a polygon of an even number of equal sides, that shall not meet the lesser circle.
Through the centre K draw the straight line BD, and from the point G, where it meets the circumference of the lesser circle, draw GA at right angles to BD, and produce it to
Boox XIL C; therefore AC touches a the circle EFGH: Then, if the
circumference BAD be bisected, and the half of it be again • 16. 3.
bisected, and so on, there must at length remain a cireum. 6 Lemma. ference less b than AD: Let this be LD; and from the point L draw LM perpendicu
DOT 1415 1 lar to BD, and produce it to N;
Alonso and join LD, DN. Therefore
H.NL LD is equal to DN;- and because LN is parallel to AC, and E
K GMD that AC touches the circle EFGH; therefore LN does not meet the circle EFGH. And much less shall the straight
с lines LD, DN, meet the circle EFGH: So that if straight lines equal to LD be applied in the circle ABCD from the point L around to N, there shall be described in the circle a polygon of an even number of equal sides not meeting the lesser circle. Whịch was to be done.
If two trapeziums ABCD, EFGH be inscribed in the circles, the centres of which are the points K, L; and if the sides AB, DC be parallel, as also EF, HG; and the other four sides AD, BC, EH, FG, be all equal to one another; but the side AB greater than EF, and DC greater than HG: the straight line KA from the centre of the circle in which the greater sides are, is greater than the straight line LE drawn from the centre to the circumference of the other circle.
If it be possible, let KA be not greater than LE; then KA must be either equal to it, or less. First, let KA be equal to LE: Therefore, because in two equal circles AD,
BC, in the one, are equal to EH, FG in the other, the cir• 28. 3. cumferences AD, BC, are equal a to the circumferences EH,
FG; but because the straight lines AB, DC are respectively greater than EF, GH, the circumferences AB, DC are greater than EF, HG: Therefore the whole circumference ABCD is greater than the whole EFGH: but it is also
equal to it, which is impossible: Therefore the straight Boos XII. line KA is not equal to LE.
But let KA be less than LE, and make LM equal to KA, and from the centre L, and distance LM, describe the circle MNOP, meeting the straight lines LE, LF, LG, LH, in M, N, O, P; and join MN, NO, OP, PM, which are respectively parallela to and less than EF, FG, GH, HE:- 2.6. Then because EH is greater than MP, AD is greater than MP; and the circles ABCD, MNOP are equal; therefore
the circumference, AD is greater than MP; for the same reason, the circumference BC is greater than NO; and because the straight line AB is greater than EF, which is greater than MN, much more is AB greater than MN: I'herefore the circumference AB is greater than MN; and for the same reason, the circumference DC is greater than PO : Therefore the whole circumference ABCD is greater than the whole MNOP; but it is likewise equal to it, which is impossible: Therefore KA is not less than LE; nor is it equal to it; the straight line KA must therefore be greater than LE, Q. E. D.
Cor. And if there be an isosceles triangle, the sides of which are equal to AD, BC, but its base less than AB the
greater of the two sides AB, DC; the straight line KA , may, in the same manner, be demonstrated to be greater
than the straight line drawn from the centre to the cir. cumference of the circle described about the triangle.