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Book XII. GN, fo • is BA to GF, and therefore the duplicate ratio of BM

to GN, is the fame with the duplicate ratio of BA to GF. but € 4.6

the ratio of the square of BM to the square of GN, is the duplicate (ro.Def si

8 ratio of that which BM has to GN ; and the ratio of the polygon 6:20. 6. A

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ABCDE to the polygon TGHKL is the duplicates of that which
BA has to GF. therefore as the square of BM to the fquare
of GN, so is the polygon ABCDE to the polygon FGHKL.
Wherefore funilar poyguns, &c. Q. E. D.

PROP. II.

II. THEO R.

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CIR
"IRCLES are to one another as the squares of their

diameters.

Lct ABCD, ETCI be two circles, and ID, TH their diame. ters. as the touare of BD) to thu fyüale of 1H, to is the circle ALCD 10 thics circle EFGH.

For, if it be not to, the square of BD shall be to the square of FH, as the circle ALCD is to fcmc fpace either loi than the circle EFGH, or gicater than it *. Fill, let it be to a space S leis.than the circle EI GH; and in the circle EFGH delcribe the square LI CH. this square is greater than half of the circle EICH; because if througl, the points E, F, G, H, there be drawn tangents to the circle, the square EFGH is half ‘ of the square defuribed

2. 41. I.

For there is some fqrare equal to i nals; that is, to the squares of BD, the circle ABCD; It P be the hecof it. i FH and the circle ABCD it is possille and tu ihree straight lines ED, FH ard there may be a fourth proportional. P, there can be a tourth proportional, Let this be S. and in like manner are lei this te Q. therete're the qrares of o be orderstood fome things in some thele tous aight lities are proportio- , of the following Propositions.

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about the circle; and the circle is less than the square described about Book XII.
it; therefore the square EFGH is greater than half of the circle. m
Divide the circumferences EF, FG, GH, HE, each into two equal
parts in the points K, L, M, N, and join EK, KF, FL, LG, GM,
MH, HN, NE. therefore each of the triangles EKF, FLG, GMH,
HNE is greater than half of the segment of the circle it stands in;
because if straight lines touching the circle be drawn through the
points K, L, M, N, and parallelograms upon the straight lines EF,
FG, GH, HE be completed; each of the triangles EKF, FLG,
GMH, HNE Mall be the half of the parallelogram in which it is. a. 41. 1.
but every segment is less than the parallelogram in which it is.
wherefore each of the triangles EKF, FLG, GMH, HNE is greater
than half the segment of the circle which contains it, and if these
circumferences before named be divided each into two equal parts,
and their extremities be joined by straight lines, by continuing to do

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this, there will at length remain segments of the circle which toger
ther shall be less than the excess of the circle EFGH above the
space S. because, by the preceding Lemma, if from the greater of
two unequal magnitudes there be taken more than its half, and froin
the remainder more than its half, and so on, there shall at length
remain a magnitude less than the least of the proposed magnitudes.
Let then the segments EK, KF, FL, LG, GM, MH, IIN, NE be
thofe that remain and are together less than the excess of the circle
EFGH above S. therefore the rest of the circle, viz. the polygon
EKFLGMHN is greater than the space S. Describe likewise in the
circle ABCD the polygon AXBOCPDR fimilar to the polygon
EKFLGMHN. as, therefore, the square of BD is to the square of
FH, fobis the polygon AXBOCPDR to the polygon EKFLGMHN. b. s. 18,
but the square of BD is also to the square of FH, as the circle ABCD

1

c. 11. S.

2.1 4. s.

Book XII.

is to the space S. therefore as the circle ABCD is to the space S, fe
is the polygon AXBOCPDR to the polygon EKFLGMHN. but
the circle ABCD is greater than the polygon contained in it; where-
fore the space S is greater d than the polygon EKFLGMHN. but
it is likewife less, as has been demonstrated; which is impossible.
Therefore the square of BD is not to the square of FH, as the circle
ABCD is to any space less than the circle EFGH. in the fame man-
ner it may be demonstrated that neither is the square of FH to the
square of BD, as the circle EFGH is to any space less than the
circle ABCD. Nor is the square of BD to the square of FH, as the
circle ABCD is to any space greater than the circle EFGH. for, if
posible, let it be fo to T a space greater than the circle EFGH.
therefore, inversely, as the square of FH to the square of BD, so is

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the space T to the circle ABCD. but as the space + T is to the circle ABCD, so is the circle EFGH to some space, which must be lefs than the circle ABCD, because the space T is greater, by hypothesis, than the circle LFGH. therefore as the square of FH is to

† For as in the foregoing Note at manner there can be a fourth proporit was explained how it was possible tional to this other space, named T, there could be a fourth proportional and the circles ABCD, EFGH, and the to the squares of BD, FH and the circle like is to be understood in some of the ABCD, which was named S. so in like following Propositions.

the square of BD, so is the circle EFGH to a space less than the Book XII. circle ABCD, which has been demonstrated to be impoflible. therefore the square of BD is not to the square of FH, as the circle ABCD is to any space greater than the circle EFGH. and it has been demonstrated that neither is the square of BD to the square of FH, as the circle ABCD to any space less than the circle EFGH. wherefore as the square of BD to the square of FH, fo is the circle ABCD to the circle EFGHf. Circles, therefore, are, &c. Q. E. D.

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EVERY pyramid having a triangular base, may be see N.

divided into two equal and similar pyramids having triangular bases, and which are similar to the whole pyramid; and into two equal prisms which togeth are greater than half of the whole pyramid.

Let there be a pyramid of which the base is the triangle ABC and its vertex the point D. the pyramid ABCD may be divided into two equal and similar pyramids having triangular 'bases, and similar to the

D whole ; and into two equal prisms which together are greater than half of the whole pyramid.

Divide AB, BC, CA, AD, DB, DC, each into two equal parts in the points E,

KA F, G, H, K, L, and join EX, EG, GH,

L HK,KL, LH, EK, KF, FG. Because AE is equal to EB, and AH to HD, HE is parallel 'to DB. for the same reason, HK is

C parallel to AB. therefore HEEK is a pa

b. 34. I. rallelogram, and HK equal b to EB. but EB is equal to AE; therefore allo AE is B T C equal to HK. and AH is equal to HD; wherefore EA, AH are equal to KH, HD, each to each ; and the angle EAH is equal to the angle KHD; therefore the base EH is c. 29. 1.

a. 2. 6.

Because as a fourth proportional to the squares of BD, FH and the circle ABCD is posible, and that it can neither be less nor greater than the circle EFGH, it must be equal to it.

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Book XII. equal to the base KD, and the triangle AEH equal é and similar to

the triangle HKD. for the same reafon, the triangle AGH is equal d. 4. I.

and similar to the triangle HLD. and because the two straight lines
EH, HG which meet one another are parallel to KD, DL that
meet one another, and are not in the same plane with them, they
contain equale angles; therefore the angle EHG is equal to the
angle KDL. again, because EH, HG are equal to KD, DL, each
to each, and the angle EHG equal to the angle KDL ; therefore
the base EG is equal to the base KL, and the triangle EHG equal
d and similar to the triangle KDL. for the same reason, the triangle
AEG is also equal and similar to the triangle HKL. Therefore

the pyramid of which the base is the triangle AEG, and of which f. C. 11. the vertex is the point H, is equal f and

fimilar to the pyramid the base of which
is the triangle KHL, and vertex the point
D. and because HK is parallel to AB a side
of the triangle ADB, the triangle ADB is
equiangular to the triangle HDK, and
their fides are proportionals 8. therefore

K

I
the triangle ADB is fimilar to the triangle
HDK. and for the same reason, the tria
angle DBC is similar to the triangle DKL;
and the triangle ADC to the triangle HDL;
and also the triangle ABC to the triangle
AEG. but the triangle ALG is fimilar to

the triangle HKL, as before was proved, B
h. 21. 6. therefore the triangle ABC is similar k to the triangle HKL. and the

pyramid of which the base is the triangle ABC, and vertex the point i. B. 11. & D, is therefore similar i to the pyramid of which the base is the tri11.Del.it. angle HKL, and vertex the fame peint D. but the pyramid of which

the base is the triangle HKL, and vertex the point D, is similar, as has
been proved, to the pyramid the base of which is the triangle AEG,
and vertex the point H. wherefore the pyramid the base of which
is the triangle ABC, and vertex the point D, is similar to the py-
rainid of which the base is the triangle AEG and vertex H. there-
fore each of the pyramids AEGH, HKLD is similar to the whole

pyramid ABCD, and because BF is equal to FC, the parallelogram k. 41. 1.

EBFG is double k of the triangle GFC. but when there are two
prisors of the fame altitude, of which one has a parallelogram for

$. 4 6.

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