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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
ABCDE to the polygon TGHKL is the duplicates of that which
II. THEO R.
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.
about the circle; and the circle is less than the square described about Book XII.
this, there will at length remain segments of the circle which toger
c. 11. S.
2.1 4. s.
is to the space S. therefore as the circle ABCD is to the space S, fe
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.
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.
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
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
the triangle HKL, as before was proved, B
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
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
$. 4 6.