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PROPOSITION IX. THEOREM IX. HE bases (ABC & EFG), and altitudes (B D & F H), of equal pyramids, (A B C D & E F G H), having triangular bases, are reciprocally proportional, (ibat is, the bale ABC: bale EFG = altitude FH : altiinde B D), and triangular pyramids (A B C D & E F G H), of which the bases (A B C & EFG), and altitudes (B D & FH), are reciprocally proportional: are equal to one another. Hypothesis.

Thefis. 1. The pyrams. ABCDEFGH are triangular. Bafe ABC : base EFG = altitude II. The pyram. ABCD is = to the pyram. EFGH. FH: altitude B D.

Preparation
Complete the BBO & F K having the same altitude with
the pyramids ABCD & EFGH; as also the prisms
BAPNC & FELIG.

I. DEMONSTRATION.

Because the prisims PNB & Lif, have the fame base &

ECA SE the prisms P NB LIF altitude with the given pyramids A B C D & E F G H. (Prep). 1. Each prism will be uriple of its pyramid, (that is, the prisni P N B

triple of the pyramid ABCD, & the prisin L I F triple of the SP. 7. B.12, pyramid EFGH).

Cor. 1. 2. Consequently, the prism PNB is = to the prism LIF.

Ax.6. B. 1, But the ABO is double of the prism PNB, & the OFK double of the prism LIF.

P.28. B.u. 3. Therefore, the EBO is = to the EFK.

Ax.6. B. I. But the equal (BO & F K) have their bases and altitudes reciprocally proportional that is, hafe BQ : base F M = altitude FH : altitude B D). And those are each fextuple of their pyramids, (that is, the E BO is = fix pyramids A B CD, & the KF=fix pyramids EFGH. Arg. 1. & 3).

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Moreover, the base of the pyramid ABCD is the half of the base
of the GBO.
And the base of the pyramid EFGH is the half of the base >P.41. B. 1.

of the EFK.
4. Confequently, base A B C: base EFG = alt. FH: alt. B D. SP.15. B. 5.

P.11. B. 5.
Which was to be demonstrated.
Hypothefis.

Thesis.
1. The pyramids ABCD & EFGH are triangular. The triangular pyramid ABCD is =
11. Baje ABC : baje EFG=alt. FH : alt. BD. to the triangular pyramid EFGH.

II. DEMONSTRATION.

BECAUSE

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ECAUSE the A ABC:A EFG = FH: BD. (Hyp. 2). And the pgr. B Q is double of the A ABC, the pgr. F M double of the AÉFG

P. 41. B. 1. i. It follows, that the pgr. B Q: pgr. FM= FH: B D.

P.is. B. 5. But 6 B O has for base the pgr. B Q, & for alt. B D.

@ & }(Prep.). 2. Contequently, the EBO is = to the FK

P.34. B.11.
But the E BO & F K are each double of the prisms PNB &
LIF

P.IS. B.11.
And those prisors PNB&LIF are each triple of their pyramids SP. 7. B.12-

A B C D & E F G H. 3. Therefore, the triangular pyramid A B C D is = to the triangular pyramid E F G H

Ax.7. B. 1. Which was to be demonstrated. COROLLA R r. QUAL polygon pyramids have their bases and altitudes reciprocally proportional; & polygon pyramids whose bases & altitudes are reciprocally proportional: are equal.

Cor.1.

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PROPOSITION X. THEOREM X VERY cone (BRC) is the third part of the cylinder (HGFE ABDC) which has the same base, (BD C A) and the same altitude (BH) with it. Hypothefis.

Thesis. The cone BRC, & the cylinder HF ADC, The cone B RC is equal to the third bave the fame base BÓCA, & the same part of the cylinder HFCABD. altitude B H.

DEMONSTRATION,

If not,

The cone will be < or > the third part of the cylinder, by
a part = 2

1 I. Suppofition.
Let the third part of the cylinder HC be = cone B RC
+ 2.

1. Preparation.
N the base A BDC of the cone & cylinder, describe the o

, ABD C.

P. 6. B. 4. 2. About the same base describe the OPOQS.

P.
7:
B.

4. 3. Upon those squares erect two 8, the first EFHBC, upon

the inscribed Q, & the second, on the circumscribed I, which will touch the superior base with its plle. planes, in the points H, G, F, & E, * having the fame altitude with the cylinder, & che cone.

S s.
We bave omitted a part of the preparation in the figure to avoid confufion.

1.

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4. Bilect the arches ATC, CID, D 6B, & B a A, in T,2,b,& a. P.30. B. 3. 5. Draw A T, & TC, &c.

Poš.1. B. i. 6. Thro' the point T, draw the tangent ITK, which will cut BA & P.17. B. 3.

D C produced, in the points I & K & complete the pgr. A K. 7. Upon

the pgr. AK, erect the 6 ALFK, & upon the , AIT, TAC, & TCK the prisins E TI, ETF, & T FK, having all

the fame altitude with the cylinder & cone. B. Do the same with respect to the other segments A a B, B b B, &c. Because the OPOQS is described about, & the o

BD C A described in the O. (Prep. 1. & 2). 1. The OPOQ S is double of the OBD CA.

P.47. B. 1. And the described upon those squares having the same altitude, (Prep. 3). 2. Therefore, the Eupon POQS is double of the upon BDCA. P.32. B.11. But the Eupon POQS is > the given cylinder.

Ax.8. B. 1. 3. Therefore, the EP upon BDCA is the half of the same cylinder. P.19. B. 5. And since the ATAC is the half of the

pgr.
A K.

P.41. B. i. 4. The prism ETF, described upon this ATAC, will be the (P.28. B.11. half of the Eupon the pgr. A K.

P.34. B.11. The 6 described upon the pgr. A K is > the element of the (Rem.1.Cor.3

. cylinder, which has for base the segment A TC.

Ax.8. B. i. 5. Consequently, prison ETF described upon A TAC is half of

the element of the cylinder which has for base segment ATC. P.19. B. 5. 6. Likewise, all the other prisms described after the fame manner, will

be > the half of the corresponding parts or elements of the cylinder. Therefore, there may be taken from the whole cylinder more than the half, (viz. the upon the E BDCA), & from those remaining elements (viz. CFEAT, &c.) more than the balf; (viz. the prisms & TF, &c.), & so on.

B.12.

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7. Until there remains several elements of the cylinder which together will be < 2..

Lem. B.12. But the cylinder is = to three times the cone B RC+2. (Sup.). Therefore, if from the whole cylinder be taken those elements (Arg. 7.).

And from three times the cone B Å C+2, the magnitude Z. 8. The remaining prism (viz. that which has for base the polygon A a B 6 DICT) will be > the triple of the cone.

Ax.4. B. 1, But this prism is the triple of the pyramid of the fame base & alti- SP. 7. tude (viz. of the pyramid T A a B 6 DICTR).

Cor. 2. 9. Consequently, the pyramid AB DCR is the given cone. Ax.7. B. 1.

But the base of the cone is the o in which this polygon A BDC is inscribed, (& which is consequently > this polygon), & this cone

has the same altitude with the pyramid. 10. Therefore, the part is the whole. 11. Which is impossible.

Ax.8, B. 1. 12.Consequently, the cone is nọt the third part of the cylinder,

II. Supposition.
Let the cone be > the third part

of the cylinder by the mgn.
2, that is, the cone = the third part of the cylinder + 2.

II. Preparation.
Divide the given cone into pyramids, in the same manner.

that the cylinder was divided in the firfi supposition,
F from the given cone be taken the pyramid which has for base the
O ABD Č, (which is greater than the half of the whole base of
the given cone, being the half of the circumscribed O, Arg. 1. &
this being > the base of the cone, Ax. 8. B. 1.), & from the
remaining segments, the pyramids corresponding to those segments,

(as has been done in the cylinder Arg. 7.). 13. There will remain several elements of the cone which together will be < 2.

Lem. B.12. Therefore, if from the cone those elements be taken which are

2, & from the cylinder + 2, the magnitude Z. 14. The remainder, viz. the pyramid Aa Bb DICTR is = to the third part of the cylinder.

Ax.5. B. 1. But the pyr. A a BbDICTR is = to the third part of the prism, SP. 3. B.12.

. , 15. Therefore, the given cylinder, is = to this prism.

Ax.6. B. I. But the base of the given cylinder is > the base of the prism since

this second is inscribed in the first. (1. Prep. 4. & 5). 16. Therefore, the part is = to the whole. 17. Which is impossible.

Ax.8. B. 1. 18. Therefore, the third part of the cylinder is not < the cone.

And it has been demonftrated (Arg. 12.), that the third part of the

cylinder is not > the cone. 19. Therefore, the cone is the third part of the cylinder of the fame base & altitude.

Which was to be demonstrated.

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