PROPOSITION XL. THEOREM XXXV.

F two triangular prifms (FL & E C) have the fame altitude (LI & AE), and the base of one (as CL) is a parallelogram (FI), and the base of the other (EC) a triangle (A B C): if the parallelogram be double of the triangle, the first prifm (L F) will be equal to the fecond (E C).

BECAUSE

DEMONSTRATION.

ECAUSE the pgr. F I, bafe of the prifm FL, is double of the
AABC, bafe of the prifm EC (Hyp 2. & 3).

And the pgr. BO is alfo double of the ▲ A B C.

2. The BD is to the

to the altitude A F (Hyp. 1),

1. The pgr. FI is to the pgr. B O.
Moreover, the altitude LT being
NI.
-
The given prifm LF is the half of the
And the prifm EC is the half of the
3. Confequently, the prifm FL is to the

P.41. B. 1.

P.31. B.11.

N D.

B D.

P.28. B.11.

prifm E C.

Ax. 7. B. 1.

Which was to be demonftrated.

SIMILAR

THEOREM I

IMILAR polygons (ABCDE & FGHIK), infcribed in circles are to one another as the fquares of their diameters (EL & G M).

Hypothefis.

Thefis.

1. The polygons ABCDEFGHIK. Polyg. ACE: polyg. FIH=the O of the diam. ELOof the diam. G M, or as diam. EL2: diam. G M2.

are .

II. They are infcribed in circles.

BECA

Preparation.

L. In the OA CD, draw AL, & B E, also diam. E L.

}

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

DEMONSTRATION.

ECAUSE the polygons ABCDE & GFKIH are c (Hyp 1). And the VA or EA B is to VGFK, & AE: AB-FG:FK (D. 1. B. 6).

1. The AABE is equiangular with the AFG K. 2. Wherefore, A ABE is a to AGFK, &

= d.

P. 6. B. 6.

ab, also

c

But ELA is VEBA, or a, & VG MFG KF or

fecond (Arg. 3. & 4).

5.

6. Therefore,

▲ ALE & GFM, the two VELA

to the two VGMF & G F M of the

The third AEL of the AEAL will be to the third
FGM of the AF MG.

7. And alternando

But AE & G F are homologous fides of the polygons ABD & FHK. Befides, EL & G M are the diameters of the O in which thofe polygons are infcribed.

=

8. Wherefore, polyg. A B C DE: polyg. FKIHG EL2: G M2. P.22. B. 1.

Which was to be demonftrated.

IF 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 fo on there fhall at length remain a magnitude (K B), less than the leaft (C), of the propofed magnitudes.

Preparation.

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

4.

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

2. From A B, take a part H A > the half of A B.
3. From the remainder H B, take HK > the half of H B.
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.

BECAUS

ECAUSE the magnitude EI is a multiple greater than twice

the least magnitude C (Prep. 1).

If there be taken from it a magnitude G I = C.

1. The remainder E G will be

the half of E I.

But EI is > AB (Prep. 1). 2. Confequently, the half of EI is 3. Therefore, GE will be much But HB is the half of AB 4. Much more then G E is > H B. 5. Therefore, EF, the half of E G, is And KB is the half of H B

6. Confequently, EF is > KB.

the half of H B.

(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 demonstrated.

CIRCLE

L

IRCLES (A F D & ILP), are to one another as the fquares of their diameters (AE & IN).

Hypothefis.

In the circles AFD & ILP there has been drawn the diameters AE & IN.

If not,

Thefis.

OAFD: OILPAE2: IN2.

DEMONSTRATION.

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

I. Suppofition.

Let T be <OILP by the fpace V. that is, T + V
=OILP.

P. 6. B. 4.

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

P.30. B. 3.

3. Draw the lines IK, KL, LM, MN, NO, OP, PQ
& 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 as to the polygon of
the OIL P.

But the infcribed

will be > the half of the

ILP.

L N P is to half of the circumfcribed

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

of the diameter=LI+LN = 2 OLI).
LIPN is the half of the

Ax.8. B. 1.

P.19. B. 5.

P.47. B. 1.

2. Therefore, the The rgle. SI is

IL P.

Ax. 1. B. 1.

3. Confequently, the LKI.

P.19. B. 5.

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

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

6. Wherefore, the fum of all thofe triangles will be > the fum of the
half of all thofe fegments.

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

It will be proved after the fame manner.

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

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

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

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

Therefore, taking those fegments L K I, &c. from the ILP.
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. I LOQ=□ of AE of IN.

P41. B. i. P.19. B. 5.

Lem. B. 12.

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