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PRO P. XIII.
then, as one cylinder is to the other cylinder, fo is the axis of the ofte to the axis of the other.
Let the cylinder AD be divided by the plain GH, parallel to the opposite plains AB, CD, and meeting the axis EF in the point K; then, as the cylinder BG is to the cylinder GD, fo is the axis EK to KF.
For, let the axis EF be produced both ways to L and M; let EL be taken any multiple of EK; and FM any multiple of FK; through the points L, N, X, M, draw plains parallel to AB, CD, and with the centers L, N, X, M, draw the circles OP, RS, TY, VQ, each equal to AB; and compleat the cylinders PR, RB, DT, TQ; then, because the axis LŃ, NE, EK, are equal, the cylinders PR, RB, BG, are equal a. For the same reason, the cylinders HC, DT, TQ, are equal; therefore the cylinder PG is the same multiple of the cylinder BG, that the axis LK is of EK. For the fame reason, the cylinder GQ is the same multiple of GD that KM is of KF; therefore, if KL is equal to KM, PG will be equal to
GQ; if greater, greater, and, if less, less. Therefore, AH is b def. 5. 5. to GD as EK is to HF b. Wherefore, &c.
ONES and cylinders, confiituted upon equal bases, are to one another as their altitudes.
Let the cylinders EB, FD, stand upon equal bases AB, CD, then the cylinder EB is to the cylinder FD, as the altitude GH is to the altitude KL.
For, produce the axis KL to the point N, and put LN equal to GH, and let the cylinder CM be drawn about the axis LN; then the cylinders FB, CM, are to each other as their bafes "; but their bafes are equal; therefore the cylinders EB, CM, are equal; but the cylinders CM, FD, are as their axes LN, KL"; but the cylinders CM, EB, are cqual; and their axes GH, LN, likewife equal; therefore the cylinder EB is to the cylinder ID
as the axis GH to the axis KL ; but, as the cylinder EB is to $15. 5. and the cylinder FD, so is the cone ATG to the cone CDK'; there
fore, as the axis GH is to KL, 10 is the cone ABG to CDK ; and so the cylinder EB to FD. Wherefore, &c.
PRO P. XV.
THE O R.
HE bases and altitudes of equal cones and cylinders are reci
procally proportional, and cones and cylinders whose base and altitudes are reciprocally proportional, are equal to one another.
Let the circles ABCD, EFGH, be the bases of the equal cones and cylinders, AC, EG, their diameters, and KL, NN, their axes ; compleat the cylinders AX, EO; then, as ABCD is to EFGH, fo is the altitude MN to KL.
For the altitudes KL, MN, are either equal or not. If equal, the cylinders AX, EO; are likewise equal; then the bases ABCD, EFGH, are equal“; therefore the bales ABCD, EFGH, á 11. are to one another as their altitudes: But, if the altitudes KL, MN, are not equal, let one of the'n, as MN, beth greater, and cut off PM equal to LK, and let th plain YS, paraliei to the opposite plains ; cut the cylinder to in the point P, and compleat the cylinder ES; then thi. cylinde: AX is to the cylinder ES as the cylinder BO is to the cylinder ES b: but the cylinder b 7.5. AX is to the cylinder ds rhe bale ALCD to the bale EFGH“; and, as the cylinder 1.0 is to tie cyıinder ES, fo is the altitude MN to the altitude MPc; therciore the base ABCD C 13. is to the bale EFGH as the altitude MN is to the altitude KL; therefore the bates and altitudes of the cylinders AX, LO, are reciprocally proportional.
And, if the bales and altitudes of the cylinders AX, EO, are reciprocally proportional, then the cylinders are equal; for, the farne conitruction reinaining, the base ABCD is to the base EFGH, as the altitude MN is to the altitude KL; and the al. titudes KL, MP, are equal; therefore the base ABCD is to the base EFGH as the cylinder AX is to the cylinder ES ", and the altitude MN is to the altitude MP as the cylinder £O is to ES“; therefore the cylinder AX is to ES as EO is to ES ; therefore the cylinder AX is equal to Loe, and, because cones are e 2. s. one third of the cylinder of the fame bafe and altitude, and i 1o. parts have the same proportions as their like multiples 8; there fore the base and altitudes of equal cones and cylinders are reciprocally proportional. And, &c.
g 15. 5.
PRO P. XVI.
TWO circles about the same center, to inscribe
in the greater a polygon of equal fides, even in number, that shall not touch the lefjer circle.
a 16. 3.
Let ABCD, EFGH, be two given circles, about the same center K; it is required to inscribe a polygon in the circle ABCD, of equal fides, even in number, that shall not touch the leffer circle EFGH.
Through the center K draw the right line BD, and through the point G draw AG at right angles to BD : produce AG to C; then AC is a tangent to the circle EFGH, in the point G'; bifect the circumference BAD, and, again, the half thereof, and doing this, till a circumference is found less than AD, which let be LD; draw LM perpendicular to BD, and produce it to N; join LD, ND. Now, because LN is parallel to AC, the tangent of the circle EFGH, LN will not touch the circle EFGH, and, much less, the lines LD, ND; and, if right lines be applied in the circle, each equal to LD, there will be a polygon inscribed in the circle ABCD, of equal fides, and even in number, that will not touch the lesser circle EFGH; which was to be done.
c 29. 3.
PRO P. XVII. PROB.
To describe a solid polyhedron in the greater of two.spheres, har
ving the same center, which shall not touch the superficies of the leser sphere.
Let two spheres be supposed about the center A, it is required to describe a solid polyhedron in the greater sphere, not touching the superficies of the lefser sphere.
Let the sphere be cut by lome plain passing through the cena def
. 14. ter, then the sections will be circles a, and the circle described b 15: 3, by the half section will be a great circle b; which let be BEDC;
and FGH that of the lesser ; and BD, CE, two of their diameters, drawn at right angles to each other; let BD meet the lesser circle in the point G ; and draw GL a tangent to the lefser circle in the point G; and join AL: In the greater circle BEDC inscribe a polygon that will not touch the leffer circle FGH"; let the sides of the polygon, in the quadrant BE, be the right
lines BK, KL, LM, ME, such that each will subtend a less arch Book XII than a line equal to the tangent G., then the right linBK, KL, LM, ME, will each be less than the tangent GL ; and produce the lines joining the points K, A, to N; and from th.: point A raise AX perpendicular to the plain of the circle BLDC, meeting the superficies of the sphere in X d. Let plains be drawn through d 12. 11. AX, BD, and AX, KN, which will make circles in the superficies of the sphere, and let BXD, KXN, be semicircles on the diameters BD, KN; then, becaule XA is perpendicular to the plain of the circle BEDC, the semicircles BX1), KXN, are perpendicular to the same plaino; but the semicircles BED, e 18. 11. BXD, KXN, are equal, for they stand upon equal diameters BD, KN, their quadrants BE, BX, KX, full likewise be equal ; therefore, as many fides of the polygon as are in the quadrant BE, so many equal files may be in the quadrants BX, KX; let these sides be BO, OP, PR, RX, KS, ST, TY, YX; and join SO, TP, YR; from the points O, S, draw the perpendiculars OV, SQ, which will fai on BD, KN, the common fection of the plain ; join VQ; then, since the equal cir- f 38. ir. cumferences BO, SK, are taken in the equal semicircles BXD, KXN; and, because OV, SQ, are drawn perpendiculars from them, they are equal ; as also, BV, KQ; but BA, KA, are equal; therefore AV is to VB as AQ is to OK ; therefore VQ is parallel to BK , and OV is parallel to SQ"; but it is proved e- & %. 6. qual ; therefore ZV, SO, are equal and paralleli; therefore OS,
i 33. 1. BK, are parallel b; but, because BK is parallel to VQ, and AB * 9. 11. equal to AK, AB is to BK as AV is to VO 8; and altern. AB is to AV as BK is to VK; but AB is greater than AV; therefore BK is greater than VK; but VK is equal to OS; therefore BK is greater than OS ; join BO, KS, then OBKS is a quadrilateral figure in one plain!. For the same reason, each of the quadrilateral figures SOPT, TPRY, and triangle YRS, are each in one plain ; therefore, if from the points O, S, P, T, K, Y, to the point A, right lines are supposed drawn, will constitute a polyhedrous figure within the circumferences BX, KX, consisting of pyramids, whose bases are KBOS, SOPT, TPRY, YRX, and vertex the point A ; and if, in the fame manner, pyramids be constructed on the sides KL, LM, ME, and on the other three quadrants and opposite hemisphere, there will be constructed a polyhedrous figure described in the sphere, composed of pyramids whofe bafes are equal and similar to the forefaid quadrilateral figures, and triangle YRX, and vertex the point A.
But the polyhedron does not touch the fuperficies of the sphere in which the circle FGH is. For, because the quadrilateral figure KBSO is in one plain, and from the point A be drawn a right line AZ perpendicular to the plain m, it will be at right m 11. 15.
11. O 47. I.
Book XII angles to all the right lines drawn in that plain "; join BZ,
V ZK, then AZ will be perpendicular to BZ, ZK. But the 3. squares of AK, AB, are equal, and the squares of AZ, ZB, are
equal to the square of AB, and the squares of AZ, ZK, are equal to the iquare of AK "; therefore the squares of AZ, ZB, are equal to the squares of AZ, ZK. Take the common square of AZ from both, then the squares of EZ, ZK, are equal ; that is, BZ equal to ZK. In like manner, lines drawn from Z to the points O, S, may be proved equal to BZ, ZK; therefore a circle described about the center Z, with either of these distances, will pass through the points O, S, K, B; and, because KBSO is a quadrilateral figure inscribed in a circle, and OB, “K, KS, are equal, and OS less than BK, the angle BZK will be obtule ; therefore BK is greater than BZ; but GL is greater than KB, and therefore much greater than BZ; and the squares of AG, GL, are equal to the square of AL, AB, or AK; therefore the squares of BZ, ZA, are equal to the square of AL; and the squares of AZ, ZB, are equal to the squares of AG, GL; but the square of GL was proved greater than the quare of BZ; therefore the square of AZ is greater than the square of AG; that is, the right line AZ greater than AG; but AZ is perpendicular to one of the bases of the polyhedron ; and AG reaches the superficies of the lefser sphere; therefore the polyhedron does not touch the superficies of the lesser sphere. Wherefore, &c.
Cor. If a folid polyhedron is inscribed in another sphere, fimilar to that in BCDÉ, they shall be to one another in the tripli. cate ratio of the squares of their diameters; for, the folids being divided into pyramids, equal in number, and of the same order,
they will be fimilar; and therefore to one another in the tripli6 12.
cate ratio of their homologous fides P; that is, as AB drawn from the center of the sphere BCDE, to the semidiameter of the other spheret; but the femidiameters of spheres are as their diameters; and one of the antecedents is to one of the confequents as all the antecedents to all the confequents; therefore, the polyhedron in the one fphere is to the fimilar polyhedron in the other sphere, in the triplicate ratio of their diameters.
r 12. 5.
PRO P. XVIII. THEO R.
PHERES are to one another in the triplicate ratio of their