LEM M A.

IF from the greater (AB), of two unequal magnitudes (A B & C), there

be taken more than its half (viz. A H), and from the remainder (H B) more than its half (viz. H K), and fo on there fhall at length remain a magnitude (K B), less than the least (C), of the propofed magnitudes.

Preparation.

1. Take a multiple EI of the leaft C, which may surpass
A B, & be > 2 C.

Pof.1. B. 5.
Pof.2. B. 5.

2. From A B, take a part HA > the half of A B.
3. From the remainder H B, take HK > the half of H B.
4. Continue to take more than the half from thofe fuccef-
five remainders, until the number of times, be equal to the
number of times, that C is contained in its multiple E I. Pof.2. B. 5.

DEMONSTRATION.

BECAUSE the magnitude E I is a multiple greater than twice

the leaft magnitude C (Prep. 1).

If there be taken from it a magnitude GI C. the half of EI.

1. The remainder E G will be

But EI is AB (Prep. 1).

2. Confequently, the half of EI is

3. Therefore, GE will be much

the half of A B.
the half of A B.

P.19. B. 5.

But H B is the half of A B (Prep. 2).

4. Much more then G E is > H B.

5. Therefore, EF, the half of E G, is > the half of H B.

And KB is the half of H B

6. Confequently, EF is > KB.

(Prep. 3).

And as the fame reafoning may be continued until a part (E F) of the multiple of the magnitude C be attained, which will be equal to C (Prep. 4).

7. It follows, that the magnitude C will be > the remaining part (KB) of the greater A B.

Which was to be demonftrated.

PROPOSITION II. THEOREM II.

CIRCLES (AFD & IL P), are to one another as the squares of

their diameters (AE & IN).

A E2 is to IN2 as the AFD is to a space T (which
is <or the OIL P).

P. 6. B. 4.

2. Divide the arches IL, LN, NP, & PI into two equal
parts in the points K, M, O, & Q.

P.30. B. 3.

3. Draw the lines I K, KL, LM, MN, NO, O P, P Q
& Q I.

Pof.1. B. 1.

4. Thro' the point K, draw S R plle. to LI.

P.31. B. 1.

5. Produce NL & PI to R & S; which will form the rgle.
SRIL.

6. Inscribe in the ADF a polygon to the polygon of

L N P is to half of the circumfcribed

1. The half of this

will be > the half of the

But the infcribed

of the

diameter=OLI+OLN=2OLI).

(the fide of the circumfcribed being to the diameter, & the

2. Therefore, the LIPN is the half of the Í LP.

Ax.8. B. 1.

P.19. B. 5.

P.47. B. 1.

Ax.1. B, 1.

the half of the fegment

The rgle. SI is the fegment L KI (Prep. 5. & Ax. 8. B. 1). 3. Confequently, the half of the rgle. SI is

LKI.

The ALKI is to half of the rgle. S I.

4. Therefore, the ALKI is the half of the segment L K I.
5. It may be proved after the fame manner, that all the ALM N,
NOP, &c. are each > the half of the fegment in which it is
placed.

6. Wherefore, the fum of all those triangles will be > the fum of the
half of all those segments.

Continuing to divide the fegments KI, IL, &c. as also the segments arrifing from those divifions.

It will be proved after the fame manner.

7. That the triangles formed by the ftraight lines drawn in those segments, are together the half of the fegments in which those triangles are placed.

Therefore, if from the ILP be taken more than its half, viz. the ILNP, & from the remaining fegments (L KI, IQP, &c.) be taken more than the half, & fo on.

8. There will at length remain fegments which together, will be < V.

But the ILP is T+ V. (1: Sup.).

Therefore, taking thofe fegments L K I, &c. from the OIL P.
And the space V, from TV (which is thofe fegments).
9. The remainder, viz. the polygon IKLMNOPQ will be > T.
But the polyg. ADFK: polyg. ILOQ=□ of AE of IN.

Lem. B.12.

Ax.5. B. 1.
P. 1. B.12.

And the of AEO of INACEG: T. (Sup.). 10. Therefore, the polyg. ADFH: polyg. ILOQ=ACEG: T. But the polygon ADF H is <ACEG. 11.Confequently, the polygon I LOQ is < T. But the polygon ILOQ is > T. (Arg. 9).

12. Therefore, T will be > &< the polyg. ILOQ (Arg. 9. & 11). 13.Which is impoffible.

14.Therefore, T is not < OIL P.

15. From whence it follows, that the of the diameter (A E) of a
(ACEG), is not to the of the diameter (IN) of another
(ILP), as the firft

(ACEG) to a space < the fecond

II. Suppofition.

Let the space T be > the circle IL P.

II. Preparation.

Take a space V, fuch that
TOACEGILP: V.

(ILP).

BECAUSE the of AE: of IN=ACEG: T.

16.Invertendo T: O ACE G=□ of IN: □ of A E.
But TO ACEG OILP V. (II. Prep.).
Moreover, T is > OIL P. (II. Sup.).
17.Confequently, the ACEG is alfo V.

Befides TACEG of INO of AE (Arg. 16).
TOACEGO ILP V. (II. Ptep.).

And

18. Therefore, the of IN: of AE OILP: V. But VO ACE G. (Arg. 17).

And it has been demonftrated (Arg.15), that the of the diameter (IN) of a (IL P), is not to the of the diameter of another O(ACEG); as the first

O(ACEG).

19.Confequently, V is not the 20. Therefore, T is not

the

Therefore, the space T being (Arg. 14. & 19).

21.T will be to this O IL P.

(ILP) to a space < the second

IL P.
IL P.

neither < nor > the IL P,

P.11. B. 5.

Ax.8. B. 1.
P.14. B. 5

P. 4. B. 5.
Cor.

P.14. B. 5.

P.11. B. 5.

22.Confequently, the OA CEG: OILP of AE: □ of IN. P. 7. B. 1.

CIRCLES

Which was to be demonftrated.

COROLLA RY.

IRCLES are to one another as the polygons infcribed in them (P. 1. B. 12. & P. 11. B. 5).

PROPOSITION III. THEOREM III.

EVERY pyramid (ABCD) having a triangular bafe (A CD), may

be divided into two equal and fimilar prifms, (I DE FLG & GLFHCE), and into two equal and fimilar pyramids, (LGIA & L F H B), which are fimilar to the whole pyramid; and the two prifms together are greater than half of the whole pyramid (A B C D).

Hypothefis.
ABCD is a pyramid whofe bafe

ADC is a A.

Thefis.

I. Preparation.

1. Cut all the fides of the pyramid A B C D into two equal
parts, in the points L, F, H, E, G, & 1.

2.

Draw the lines L F, F H, FE, GE, GI & IL, also
LG, & LH.

BECAUSE DEMONSTRATION; BC

ECAUSE in the ABCD the fides B D & B C are divided into two equal parts in the points F & H (Prep. 1).

BHHCBF: DF.

3. Confequently, FH is plle. to DC. 3. Likewife, FE is plle. to B C.

4. Therefore,

are pgrs.

FECH is a pgr.

}

be proved after the fame manner, that LFEG & LGCH

And fince F H & H L are plle. to EC & GC. (Arg. 2. & 5). 6. The planes paffing thro' L'F H & ECG will be plle.

7. Therefore, LGECHF will be a prism.

8. Likewife, LFEDIG will be alfo a prifm.

}

P.10. B. 1.

Pof.1. B. 1.

P.19. B. 5.

P. 2. B. 6.

D.35. B. 1.

P.15. B.11.

D.13.B. 11.