If LI A ,

PROPOSITION XL. THEOREM XXXV.

M F two triangular prisms (FL & EC) have the same altitude (L I & A E), and the base of one (as C L) is a parallelogram (F I), and the base of the other (E C) a triangle (A B C): if the parallelogram be double of the triangle, the first prism (L F) will be equal to the second (E C). Hypothesis.

Thesis. 1. In the prisms F L & EC; the alt. LI The prison FL is = to the prism EC.

is = to the alt. A E. II. The base of the prism L F is a pgr. F I,

& the base of the prism EC a D'ABC. III. The pgr. F l is double of the A A BC.

Preparation

Complete the ENI & BD.

DEMONSTRATION.

Because

ECAUSE the pgr. F I, base of the prism F L, is double of the
A ABC, base of the prism EC (Hyp 2. & 3).
And the

Pgr.
BO is allo double of the A ABC.

P.41. B. 1. 1. The pgr. F I is = to the

Pgr.

во. Moreover, the altitude Ll being to the altitude A F (Hyp. 1), 2. The EBD is = to the ENI. The given prism L F is the half of the END.

P.31. B.11. And the prism E C is the half of the eBD.

P.28. B.u. 3. Consequently, the prism F L is to the prifin eć.

Ax.7. B. 1.
Which was to be demonstrated.

2

PROPOSITION 1. THEOREM I.
SIMILAI

IMILAR polygons (ABCDE & F GHI K), inscribed in circles
are to one another as the squares of their diameters (EL & G M).
Hypothesis.

Thesis. 1. The polygons À BCD E&FGHI K. Polyg. : ACE: polyg. FIH=the o are u.

of the diam. EL: 0 of the diam. G M, II. They are inscribed in circles.

or as diam. EL2 : diam. G M2.

Preparations.
1. In the O ACD, draw A L, & B E, also diam. E L.

In the FMH, draw the homologous lines FM & Pof.1. B. 1.
GK; also the diameter G M.

DEMONSTRATION.
BECAUSE the polygons ABCDE & GFKIH are o (Hyp 1).

And the A or EAB is to VGFK, & AE:ABFG:FK

(D. 1. B. 6).
1. The AA B E is equiangular with the AFG K.

P. 6. B. 6. 2. Wherefore, A A B E is o to AGFK, & VaV b, allo y c

Y d. But Y EL A is = VEBA, or a, & V GMF=VGKF or b. P.21. B. 6. 3. Consequently, VEL A is = to VGMF.

Ax.1. B. I. 4 Likewite, VEAL = VGFM.

P.31. B. 3. And, because, in the two A ALE & G FM, the two WELA & EAL of the first are = to the two Y GMF & G FM of the

second (Arg. 3. & 4).
5.

The third Ý A E L of the AEAL will be to the third
VFG M of the AFMG.

P.32. B. 1.
6. Therefore,
EL : AE = GM: GF.

P.

4
7. And alternando

EL: GM = AE: GF.
But A E & G F are homologous fides of the polygons ABD & FHK.
Besides, EL & GM are the diameters of the O in which those

polygons are inscribed.
8. Wherefore, polyg. ABCDE: polyg. FKIHG=ELP: GM2. P-22. B. 1.

Which was to be demonstrated.

B. 6. P.16. B. 5.

L E M M A.

M F from the greater (A B), 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 so on : there shall at length remain a magtitude (K B), less than the least (C), of the proposed magnitudes.

Preparation.
1. Take a multiple EI of the least C, which may furpass
AB, & be > 2 C.

Pol.1. B.
2. From A B, take a part HA > the half of A B. Pos-2. B. 5.
3. From the remainder H B, take HK > the half of H B.
4. Continue to take more than the half from those succes-

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.
E CAUSE the magnitude E I is a multiple greater than twice
the least magnitude C (Prep. 1).

If there be taken from it a magnitude GI=C. 1. The remainder E G will be the half of El.

But E I is > AB (Prep. 1). 2. Consequently, the half of EI is > the half of A B.

2.19. B. 5 3. Therefore, GE will be much > the half of A B.

But H B is < the half of AB (Prep. 2). 4. Much more then G E is > HB. 5. Therefore, EF, the half of E G, is > the half of H B.

And KB is the half of H B (Prep. 3). 6. Consequently, E F is > K B.

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

Which was to be demonstrated.

BECAUSE

CIRCLE

PROPOSITION II. THEOREM II. IRCLES (A F D & IL P), are to one another as the squares of their diameters (A E & IN). Hypothesis.

Thesis. in the circles AFDES I L P there has O AFD: OILP= AE : IN2. been drawn the diameters A E ES I N.

DEMONSTRATION.
If not,

A E2 is to I N2 as the O AF D is to a space T (which
is < or > the OIL P).

P. 6. B. 4.

1. Supposition.
Let T be < OIL P by the space V. that is, T + V
= OIL P.

1. Preparation.
1. In the OLIP describe the QILNP.
2. Divide the arches I L, L N, NP, & PI into two equal
parts in the points K, M, O, & Q.

P.30. B. 3:
3. Draw the lines I K, KL, L M, MN, NO, OP, PR
& QI.

Pof.1. B. 1. 4. Thro the point K, draw SR plle. to LI.

P.31. B. 1. 5. Produce N'L & P I to R&S; which will form the rgle.

SRIL.
6. Inscribe in the OA DF a polygon to the polygon of

the OIL P.

Becaus

ECAUSE the described about the OIL P is > the 0 itself.

Ax.8. B. 1. 1. The half of this will be > the half of the O IL P.

P.19. B. 5 But the inscribed O ILNP is = to half of the circumscribed o (the side of the circumscribed being to the diameter, & the of the diameter = OLI+OLN=2OLI).

P.47. B. 1. 2. Therefore, the LIPN is > the half of the O IL P. Axi. B. 1.

The rgle. Siis > the segment LKI (Prep. 5. & Ax. 8. B. 1). 3. Consequently, the half of the rgle. S I is the half of the segment LKI.

P.19. B. 5 The LKI is = to half of the rgle. S I.

P 41. B. i. 4. Therefore, the ALKI is > the half of the segment L KI. P.i9. B. 5.

5 5. It

may be proved after the same manner, that all the ALMN, NOP, &c. are each > the half of the segment in which it is

placed.
6. Wherefore, the sum of all those triangles will be > the sum of the

half of all those segments.
Continuing to divide the segments KI, IL, &c. as also the leg-
ments arrising from those divisions.

It will be proved after the same manner.
7. That the triangles formed by the straight lines drawn in those seg-

ments, are together > the half of the segments in which thote
triangles are placed.
Therefore, if from the OIL P be taken more than its half, viz.
the DILNP, & from the remaining segments (L KI, IQ P, &c.)

be taken more than the half, & so on.
8. There will at length remain segments which together, will be

Lem. B.12. But the O ILP is = T + V (1. Sup.). Therefore, taking those segments L KI, &c. from the OIL P.

And the space V, from T +V (which is > those segments). 9. The remainder, viz. the polygon I KLMNOPQ will be > T. Ax.5. B. 1.

But the polyg. ADFK: polyg. I LOQ=of AE: of IN. P. 1. B.12.

<V.