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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.
Complete the ENI & BD.
ECAUSE the pgr. F I, base of the prism F L, is double of the
P.41. B. 1. 1. The pgr. F I is = to the
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.
PROPOSITION 1. THEOREM I.
IMILAR polygons (ABCDE & F GHI K), inscribed in circles
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.
In the FMH, draw the homologous lines FM & Pof.1. B. 1.
And the A or EAB is to VGFK, & AE:ABFG:FK
(D. 1. B. 6).
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).
The third Ý A E L of the AEAL will be to the third
P.32. B. 1.
EL: GM = AE: GF.
polygons are inscribed.
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.
five remainders, until the number of times, be equal to the
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.
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.
A E2 is to I N2 as the O AF D is to a space T (which
P. 6. B. 4.
P.30. B. 3:
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.
the OIL P.
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
half of all those segments.
It will be proved after the same manner.
ments, are together > the half of the segments in which thote
be taken more than the half, & so on.
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.