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Con

PROPOSITION XI. THEOREM X1. ONES (E À BDF & HGKIM), and cylinders (Q R BE & STKH) of the fame altitude, are to one another as their bales Hypothesis.

Thesis. The cones EABDF & HGKIM, as like 1. Cone EFB : cone HMK = base EABD wise the cylinders QRBE & STKH : base HGKI. have the same altitude.

II. Cylinder QRBE : cylinder STKH =

base E ABD: base HGKI.

DEMONSTRATION.
If not, The cone e FB :Z (which is < or > the cone
HMK) = base E ABD : bale HGKI.

1. Suppofition.
Let Z be the cone HMK by a magnitude X, that is, let
the cone HMK = 2 + X.

IN OGHIK base of cone commenti delcribe CG HIK. P. 6. B. *

N

; 2. Divide the cone into pyramids (as in II. Sup. of P. 10.). 3. In the bases of the cones EFB & HMK, draw diam. EB & HK. 4. In the O E ABD base of the cone E F B, describe a polyg. v

to the polyg.HbGgKLIiH, & divide it as the cone H M K. ECAUSE the cone HMK has been divided into pyramids. (Prep. 2.). If those pyramids be taken from the cone (as was done in i be fore

going proposition. Arg: 13;). 1. The sum of the remaining elements will be < X.

Lem B.12. Therefore, if those elements be taken from the cone H M K, & the magnitude X from 2 + X.

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2. The remaining pyramid HbGgKLIIM will be > Z.

But those polygons inscribed in the OEABD & HGKI are r. (Prep. 4.). 3. Therefore, O AEDB: O GHIK = polyg. Cdea: polyg: SP. 2. B.12. ibg L.

Cor.
But,

O AEDB: :O GHIK= cone E FB : Z. (Sup.).
And the pyramid Dd Ee A a BCF : pyramid HbGgKLIIM
polygon Cdea: polygon i bg L.

P. 6. B.11. 4. Consequently, pyram. Dd Ee ABCF: pyram. HbGgKLIIM = cone EFB: Z.

P.II. B. 5 But the pyramid Dd Ee A a B CF is < cone E F B.

Ax.8. B. 5. Therefore, the pyramid Hb Gg KLIiMis < Z.

P.14. B, 5.
6. But this pyramid is > Z. (Arg. 2.)
7. Therefore, it will be > & < Z. (Arg. 2. & 6).
Ś. Which is impossible.
9. Therefore, the supposition of Z < the cone HMK is false.
10. Wherefore, the base of the cone E F B is not to the base of the

cone HMK (the cones having the same altitude) as the cone E FB
to a magnitude 2 < the cone HMK.

Il. Suppofition.
Let Z be the cone H M K.

11. Preparation.
Take a magnitude X such that 2 : cone E F B = cone
HMK: X.

Because

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P.11. B. $

ECAUSE Z is > the cone HMK. (1. Sup.). 11. The cone E F B is > X.

B.

5: B.

5
12. Therefore, base HGKI: base E ABD=Z: cone E F B. { Cor.
13. Consequently, base GHIK: base A EBD='cone HMK:X.

But it has been demonstrated ( Arg. 10.), that the base of a cone is
not to the base of another cone, having the same altitude, as the

first cone is to a magnitude < the fecond.
14. Therefore, X is not the cone E F B.

But X is the cone E F B. (Arg. 10.).
15. Consequently, X will be < & not this cone EFB. (Arg.11.814).
16. Which is impossible.
17. From whence it follows, that the suppofition of 2 > the cong

H MK is false.
Therefore, the magnitude Z being neither <nor > the cone

HM K. (Arg. 9. & 17.).
18. It will be = to the cone H M K.
19. Hence cone E FB : cone HMK = base E ABD : bafe HGKI. P.

Which was to be demonstrated. i.
EECAUSE the cone EFB is the third part of the cylin. QRBE
And the cone H M K is the third part of the cylin. HSTK.

P.10. B.12. 20. The cylin. QRBE : cyl. HSTK = base E ABD : base HGKI. P.15. B. s.

Which was to be demonftrated. II.

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SIMILAR

PROPOSITION XII. THEOR EM XII. IMILAR cones (BFE & LOM), and cylinders (Bab E & Lcd M) have to one another the triplicate ratio of that which the diameters (CD & I H) of their bases (B Y DEP & LTHMR), have. Hypothesis.

Thesis. The cones BFE & LOM, likewise the I. The cone BFE is to the cone LOM in the gylinders B ab E&Lcd M, are w. triplicate ratio of C D to IH; or as

CD: : IH:.
II. The cyl. B a b E is to the cyl. Lcd M, in

the triplicate ratio of CDIH; or as
CDI: IH

DEMONSTRATION.
If not,

The cone B F E is to a magnitude Z (which is < or the
cone LOM) as CD: : IH.

1. Supposition.
Let Z be < the cone L OM by the magnitude X, that is,
the cone LOM=2 + X.

1. Preparation
1. Divide the LOM into pyramids, as in the foregoing

propofition.
2. In the base of the cone B F E describe a polygon w to the

polygon of the base of the cone LOM.
3. In the two cones draw the homologous diameters I H &

CD; also the rays L N&B A.

BE CAU 5

L

P.11. B. Š.
P.16. B. 5.

ECAUSE the cone L OM has been divided into pyramids.
If those pyramids be taken from this cone (in the same manner as

in the foregoing proposition. Arg. 1.).
1. The suin of the remaining elements will be < X.

Lem. B.12. Therefore, if those elements be taken from the cone LOM, & the

part X from the magnitude 2 + X. 2. The remainder, viz. the pyramid LTGHMSRIO will be > Z. Ax.4. B. 1.

But the su cones have their axes & the diameters of their bases proportional.

D.24. B.11. And the cones B FE & LOM are N. (Hyp.). 3. Consequently, CD:HI= FA: ON But, CD:HI=CA:1 N.

P.15. B. 5. 4. Therefore, CA: IN = FA: ON. 5 And alternando CA:FA=IN: ON.

The AFAC&ION have the VCAF = to VINO. (Prep.3).
And the sides CA, AF; IN, O N about those equal angles pro-

portional. (Arg. 5.).
6. Wherefore, the AFA C is 3 to the AI QN.

D. 1. B. 6.

P. 4. B. 6. 8. Likewise, the BCA is to the ALIN. (VBAC being

= VLNI). (Prep. 3.). 9. Therefore, CA: BC= IN:IL.

P. 4. B. 6. But,

CF:CA= 10: IN. (Arg. 7.). 10. Consequently, CF:BC=10: IL.

In the ACAF & BAF, the fide C A is = to B A (D. 15. B. 1.)

A F is common, & VCAF = V BA F. (Prep. 3.). u. Therefore, the base B F is = to the base CF.

B. i. 12. Io like manner, L O is = to O I. But,

CF:BC= 01:IL. (Arg. 10.). 13. Therefore, BF:BC=LO: IL. 14 And invertendo, BC:BF=IL: OL.

P. 4. B. 5

. 15.Consequently, the three sides of the AB F C are proportional to Cor.

the three sides of the ALOI. 16.From whence it follows, that those ABFC & IOL are s. P. 5. B. 6. 17.It may be demonstrated after the same manner, that all the tri

angles which form the pyramid BDQF are to all the triangles which form the pyramid LHSO, each to each.

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P.22. B. 5.

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And as the bases of those pyramids are to polygons. (Prep. 2.). 18. The pyramid B DQF is to the pyramid LHSO.

D. 9. B.11. But those pyramids being s.

P. 8. B.12. 19. The pyramid B DQF: pyramid LHSO=CBS: IL.. But, CA: BC= IN:IL. (Arg. 9.).

? Cor. 20. Therefore invert. BC:CA = IL: IN.

P. 4. B. S.

Cor. 21. And alternando, BC:LI CA : IN.

P.16. B. 5. 22. Consequently, BC: LI = CD : IH.

SP.15. B. 5. 23. Therefore, three times the ratio of B C to LI is = to three times {P.11. B.

5. the ratio of C D to I H, that is, BC*: LI' =CD*:IH

But CB*: IL: = pyramid BDQF : pyramid LHSO. (Arg.19). 24. Consequently, pyramid BDQF : pyramid LHSO =CD* : IÅ,

P.U. B. 5: But the cone B F E:2=CD3: IH”. (Sup.).

P.11. B. 25. Therefore, the pyram. BDQF : pyram. LHSO =cone BFE :2.

52 But the pyramid B D QF being < cone B E F.

Ax.8. B. I, 26. The pyramid LHSO will be also < Z.

P.14. B. 1. But the pyramid LHS O is > Z. (Arg. 2.). 27. Consequently, the pyram. LHSO will be < &> Z. (Arg.2. & 26). 28. Which is imposible. 29. Therefore, the supposition of 2 < the cone LOM or LTG

HMSRI O is falle.

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