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greater than the cone EFGHN. First, let it have it to a less, viz. to the solid X. Make the same construction as in the preceding proposition, and it may be demonstrated the very same way as in that proposition, that the pyramid of which the base is the polygon EOFPGRHS, and vertex N, is greater than the solid X. Describe also in the circle ABCD the polygon ATBYCVDQ similar to the polygon EOFPGRHS, upon which erect a pyramid having the same vertex with the cone; and let LAQ be one of the triangles containing the pyramid upon the polygon ATBYCVDQ the vertex of which is L; and let NES be one of the triangles containing the pyramid upon the polygon

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EOFPGRHS of which the vertex is N; and join KQ, MS: because then the cone ABCDL is similar to the cone EFGHN, AC is (24. def. 11.) to EG, as the axis KL to the axis MN; and as AC to EG, so (15. 5.) is AK to EM; therefore as AK to EM, so is KL to MN; and, alternately, AK 10 KL, as EM to MN: and the right angles AKL, EMN are equal; therefore the sides about these equal angles being proportionals, the triangle AKL is similar (6.6.) to the triangle EMN. Again, because AK is to KQ, as EM to MS, and that these sides are about equal angles AKQ, EMS, because these angles are, each of them, the same part of four right angles at the centres K, M; therefore the triangle AKQ is similar (6. 6.) to the triangle EMS: and because it has been shown, that as AK to KL, so is EM to MN, and that AK is equal to KQ, and EM to MS, as QK to KL, so is SM to MN, and therefore the sides about the right angles QKL, SMN being proportionals, the triangle LKQ is similar to the triangle NMS: and because of the siinilarity of the triangles AKL, EMN, as LA is to AK, so is NE to EM: and by the similarity of the triangles AKQ, EMS, as KA to AQ, so ME to ES; ex æquali, (22. 5.) LA is to AQ as NE to ES. Again, because of the similarity of the triangles LQK, NSM; as LQ to QK, so NS to SM ; and from the similarity of the triangles KAQ, MES, as KQ to QA, so MS to SE; ex æquali, (22. 5.) LQ is to QA, as NS to SE: and it was proved that QA is to AL as SE to EN, therefore, again, ex æquali, as QL to LA, so is SN to NE; wherefore the triangles LQA, NSE, having the sides about all their angles proportionals, are equiangular (5. 6.) and similar to one another : and therefore the pyramid of which the base is the triangle AKQ and vertex L, is similar to the pyramid the base of which is the triangle EMS, and vertex N, because their solid angles are equal (B. 11.) to one another, and they are contained by the same number of similar planes: but similar pyramids which have triangular bases have to one another the triplicate (8. 12.) ratio of that which their homologous sides have ; therefore the pyramid AKQL has to the pyramid EMSN the triplicate ratio of that which AK has to EM. In the same manner, if straight lines be drawn from the points D, V, C, Y, B, T to K, and from the points H, R, G, P, F, 0 to M, and pyramids be erected upon the triangles having the same vertices with the cones, it may be demonstrated that each pyramid in the first cone has to each in the other, taking them in the same order, the triplicate ratio of that which the side AK has to the side EM; that is, which AC has to EG: but as one antecedent to its consequent, so are all the antecedents to all the consequents ; (12. 5.) therefore as the pyramid AKQL to the pyramid EMSN, so is the whole pyramid the base of which is the polygon DQATBYCV, and vertex L, to the whole pyramid of which the base is the polygon HSEOFPGR, and vertex N. Wherefore also the first of these two last named pyramids has to the other the triplicate ratio of that which AC has to EG. But by the hypothesis, the cone of which the base is the circle ABCD, and vertex L, has to the solid X, the triplicate ratio of that which AC has to EG: therefore as the cone of which the base is the circle ABCD, and vertex L, is to the solid X, so is the pyramid the base of which is the polygon DQATBYCV, and vertex L, to the pyramid the base of which is the polygon HSEOFPGR, and vertex N: but the said cone is greater than the pyramid contained in it, therefore the solid X is greater (14. 5.) than the pyramid, the base of which is the polygon ÀSEOFPGR, and vertex N ; but it is also less ; which is impossible ; therefore the cone, of which the base is the circle

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ABCD, and vertex L, has not to any solid which is less than the cone of which the base is the circle EFGH and vertex N, the triplicate ratio of that which AC has to EG. In the same manner it may be demonstrated that neither has the cone EFGHN to any solid which is less than the cone ABCDL, the triplicate ratio of that which EG has to AC. ABCDL have to any solid which is greater than the cone EFGHN, the triplicate ratio of that which AC has to EG: for, if it be possible, let it have it to a greater, viz. to the solid Z: therefore, inversely, the solid Z has to the cone ABCDL, the

Nor can the cone

triplicate ratio of that which EG has to AC: but as the solid Z is to the cone ABCDL, so is the cone EFGHN to some solid, which must be less (14. 5.) than the cone ABCDL, because the solid Z is greater than the cone EFGHN: therefore the cone EFGHN has to a solid which is less than the cone ABCDL, the triplicate ratio of that which EG has to AC, which was demonstrated to be impossible: therefore the cone ABCDL has not to any solid greater than the cone EFGHN, the triplicate ratio of that which AC has to EG; and it was demonstrated that it could not have that ratio to any solid less than the cone EFGHN: therefore the cone ABCDL has to the cone EFGHN the triplicate ratio of that which AC has to EG : but as the cone is to the cone, so (15. 5.) the cylinder to the cylinder; for every cone is the third part of the cylinder upon the same base, and of the same altitude: therefore also the cylinder has to the cylinder the triplicate ratio of that which AC has to EG. Wherefore similar cones, &c. Q. E. D.

PROP. XIII. THEOR. If a cylinder be cut by a plane, parallel to its opposite planes, or bases, it divides the cylinder into two cylinders, one of which is to the other as the axis of the first to the axis of the other.*

Let the cylinder AD be cut by the plane GH, parallel to the opposite planes AB, 0

L

P CD, meeting the axis EF in the point K, and let the line GH be the common section of the plane GH and the surface of the cy- R

s linder AD: let AEFC be the parallelogram, in any position of it, by the revolution of which about the straight line EF the cylinder AD is described, and let GK be the A E IB common section of the plane GH, and the plane AEFC: and because the parallel planes AB, GH are cut by the plane AEKG, AE, KG, their common sections G

K

H with it, are parallel (16. 11.): wherefore AK is a parallelogram, and GK equal to C

D EA the straight line from the centre of the circle AB: for the same reason each of the T X Y straight lines drawn from the point K to the line GH may be proved to be equal to those V M Q

N

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F

M

See Note.

which are drawn from the centre of the circle AB to its circum-. ference, and are therefore all equal to one another. Therefore the line GH is the circumference of a circle, (15. def. 1.) of which the centre is the point K: therefore the plane GH divides the cylinder AD into the cylinders AH, GD; for they are the same which would be described by the revolution of the parallelograms AK, GF, about the straight lines EK, KF: and it is to be shown, that the cylinder AH is to the cylinder HC, as the axis EK to the axis KF. · Produce the axis EF both ways; and take any number of straight lines EN, NL, each equal to EK ; and any number FX, XM, each equal to FK; and let planes parallel to AB, CD pass through the points L, N,

0

L P X, M; therefore the common sections of these planes with the cylinder produced are circles the centres of which are the points L, N, X, M, as was proved of the plane R N S GH: and these planes cut off the cylinders, PR, RB, DT, TQ: and because the axes LN, NE, EK are all equal, therefore the A

E B cylinders PR, RB, BG are (11. 12.) to one another as their bases ; but their bases are equal, and therefore the cylinders PR, RB, BG are equal : and because the axes LN, G

K H NE, EK are equal to one another, as also the cylinders PR, RB, BG, and that there

C

I N are as many axes as cylinders; therefore,

D whatever multiple the axis KL is of the axis

T KE, the same multiple is the cylinder PG

Y of the cylinder GB: for the same reason,

V

M M whatever multiple the axis MK is of the

Q axis KF, the same multiple is the cylinder QG of the cylinder GD: and if the axis KL be equal to the axis KM, the cylinder PG is equal to the cylinder GQ; and if the axis KL be greater than the axis KM, the cylinder PG is greater than the cylinder QG; and if less, less : since therefore there are four magnitudes, viz. the axes EK, KF, and the cylinders BG, GD, and that of the axis EK and cylinder BG, there has been taken any equimultiples whatever, viz. the axis KL and cylinder PG; and of the axis KF and cylinder GD, any equimultiples whatever, viz. the axis KŃ and cylinder GQ: and it has been demonstrated, if the axis KL be greater than the axis KM, the cylinder PG is greater than the cylinder GQ; and if equal, equal;

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