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Book XII.GN, fois BA to GF, and therefore the duplicate ratio of BM to GN,

is the same f with the duplicate ratio of BA to GF. but the ratio e. 4.6. of the square of BM to the square of GN, is the duplicate & ratio f.10.Def. 5. of that which BM has to GN; and the ratio of the polygon ABCDE

and 22. 5.

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to the polygon FGHKL is the duplicate 8 of that which BA has to GF. therefore as the square of BM to the square of GN, so is the polygon ABCDE to the polygon FGHKL. Wherefore similar polysons, &c. Q. E. D.

PROP. II. THEOR.

Sue N.

CIRCLES

Es are to one another as the squares of their dia

meters.

Let ABCD, EFGH be two circles, and BD, FH their diameters. as the square of BD to the square of FH, fo is the circle ABCD to the circle EFGH.

For, if it be not so, the square of BD shall be to the square of FH, as the circle ABCD is to some space either less than the circle EFGH, or greater than it *. First, let it be to a space S less than the circle EFGH; and in the circle EFGH describe the square EFGH. this square is greater than half of the circle EFGH; be

cause if through the points E, F, G, H, there be drawn tangents a. 41. . to the circle, the square EFGH is half of the square described a

For there is some square equal to the circle ABCD; let P be the side of it. and to three Straight lines BD, FH and P, there can be a fourth proportional, let this be Q. therefore the squares of these four Straight lines are proportio

nals; that is, to the squares of BD, FH and the circle ABCD it is possible there may be a fourth proportional. Let this be S. and in like manner are to be understood some things in some of the following Propofitions.

bout

bout 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. 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 shall be the half a 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 together shall be less than the excess of the circle EFGH above the space S. because, by the preceeding Lemma, if from the greater of two unequal magnitudes there be taken more than its half, and from 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, HN, NE be those 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 similar to the polygos EKFLGMAN, as, therefore, the square of BD is to the fytare of FH, so b is the polygon AXBOCPDR to the polygon EKFLGMEN. b. 1. 12. but the square of BD is also to the square of FH, as the circle ABCD

Book XII. is to the space S. therefore as the circle ABCD is to the space S, fo Mis the polygon AXBOCPDR to the polygon EKFLGMHN, but

C. 11.5. the circle ABCD is greater than the polygon contained in it; whered. 14. 5. fore the space S is greater than the polygon EKFLGMHN. but

it is likewise less, as has been demonstrated; which is impossible. Therefore the square of BD is not to the square of FH, as the circle ALCD is to any space less than the circle EFGH, in the same manner 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 poffible, 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 less than the circle ABCD, because the space T is greater, by hypothesis, than the circle EFGH. therefore as the square of FH is to

+ For as in the foregoing Note at * it was explained how it was possible there could be a fourth proportional to the Squares of BD, FH and the circle ABCD, which was named S. fo in like manner

there can be a fourth proportional to this other space, named T, and the circles ABCD, EFGH, and the like is to be understood in some of the following Tropofitions.

the

the square of BD, so is the circle EFGH to a space lefs than the Book XII. circie ABCD, which has been demonstrated to be impossible. 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 EFGH ț. Circles, therefore, are, &c. Q. E. D.

PROP. III. THEOR.

EVE

VERY pyramid having a triangular base, may be di. See N.

vided into two equal and similar pyramids having triangular bases, and which are similar to the whole pyramid; and into two equal prifins which together 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 iato two equal and similar pyramids having triangular bases, and similar to the whole; D. and into two equal prisms which together are greater than half of the whole pyramid. Divide AB, BC, CA, AD, DB, DC,

H each into two equal parts in the points E, F, G, H, K, L, and join EH, EG, GH, K

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

a. 2. 6. parallel to AB. therefore HEBK is a pa

E rallelogram, and HK equal b to EB. but

b. 34. I. EB is equal to AE; therefore also AE is B F С 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.

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

equal

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

triangle HKD. for the same reason, the triangle AGH is equal and d. 4. 1. fimilar to the triangle HLD. and because the two straight lines EH,

HG which meet one another are parallel to KD, DL that meet one 6.30. 11. another, and are not in the fame 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 4 and fimilar 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 the vertex is the f. C. 1. point H, is equal f and similar to the py

ramid the base of which is the triangle D
KHL, and vertex the point D. and be-
cause HK is parallel to AB a side of the
triangle ADB, the triangle ADB is equi-
angular to the triangle HDK, and their

H 9. 4. 6.

fides are proportionals 8. therefore the
triangle ADB is similar to the triangle K

L
HDK. and for the same reason, the tri-
angle DBC is similar to the triangle DKL;
and the triangle ADC to the triangle
HDL; and also the triangle ABC to the

EX
triangle AEG. but the triangle AEG is
similar to the triangle HKL, as before B

was proved, therefore the triangle ABC is h. 21.6. fimilar h to the triangle HKL. and the pyramid of which the base is i. B. 11. & the triangle ABC, and vertex the point D, is therefore similar i to 51. Def.12. the pyramid of which the base is the triangle HKL, and vertex the

same point 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 pyramid of which the base is the triangle AEG and vertex H. therefore each of the pyramids

AEGH, HRLD is similar to the whole pyramid ABCD. and be. k. 41... cause BF is equal to FC, the parallelogram EBFG is double k of

the triangle GFC. but when there are two prisms of the fame altitude, of which one has a parallelogram for its base, and the other

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