Sidebilder
PDF
ePub

square EFGH is half a of the square described about the circle; B. XII. and the circle is less than the square described about it; there. fore the square EFGH is greater than half of the circle. Divide a 41. 1. 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: but every segment is less than the parallelogram in which it is: wherefore each of the triang'es 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 this, there will at length re

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][ocr errors][merged small]

main segments of the circle which, together, 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 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 polygon EKFLGMHN: as, therefore, the square of BD is to the square of FH, so b is the polygon AXBOCPDR to the b 1. 12. polygon EKFLGMHN: but the square of BD is also to the square of FH, as the circle ABCD is to the space S: therefore

B. XIL as the circle ABCD is to the space S, so is the polygon w AXBOCPDR to the polygon EKFLGMHN: but the circle c 11.5. ABCD is greater than the polygon contained in it: wherefore, d 14. 5. the space S is greater d 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 ABCD 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 possible, let it be so to T, a space greater than the circle EFGH: therefore, inversely, as the square of FH to the square of BD, so is the space T to the circle ABCD.

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small]

But as the space † T is to the circle ABCD, so is the circle
EFGH to some space, which must be less d than the circle
ABCD, because the space T is greater, by hypothesis, than the
circle EFGH. Therefore, as the square of FH is to the square

† 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. So, 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 propositions.

of BD, so is the circle EFCH to a space less than the circle B. XIL 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 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, so is the circle ABCD to the circle EFGHt. Circles, therefore, are, &c. Q. E. D.

PROP. III. THEOR.

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 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 into 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, each into two equal parts in the points E, F, G, H, K, L, and join EH, EG, GH, HK, KL, LH, EK, KF, FG. Because AE is equal to EB, and AH to HD, HE is parallel a to DB:

K

L a 2. 6. for the same reason, HK is parallel to AB: therefore HEBK is a parallelogram, and HK equal b to EB: but EB is equal to AE; there.

LA

b 34. 1. fore also AE is 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

c 29.1. EH is equal to the base KD, and the triangle BF

с

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 EFGH, it must be equal to it.

D

B. XII. AEH equal d and similar to the triangle HKD: for the same

reason, the triangle AGH is equal and similar to the triangle d 4. 1. HLD: and because ihe two straight lines EH, HG which meet

one another are parallel to KD, DL that meet one another, and e 10. 11. are not in the same plane with them, they contain equal e 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 iriangle AEG is also equal and similar to the triangle HKL. Therefore the py

ramid of which the base is the triangle AEG, and of which the fC. 11. vertex is the point H, is equal f and similar to the pyramid the

base of which is the triangle HKL, and ver-
tex the point D: and because HK is parallel
to AB a side of the triangle ADB, the tri-

angle ADB is equiangular to the triangle 8 4. 6. HDK, and their sides are proportionals 8 :

therefore the triangle ADB is similar to
the triangle HDK: and for the same rea-
son, the triangle DBC is similar to the tri-

K

L angle DKL; and the triangle ADC to the triangle HDL; and also the triangle ABC to the triangle AEG: but the triangle AEG is similar to the triangle HKL, as before

was proved; therefore the triangle ABC is h 21. 6. similar h to the triangle HKL. And the

pyramid of which the base is the triangle

ABC, and vertex the point D, is therefore B F с i B. 11. & similari to the pyramid of which the base 11. def. is the triangle HKL, and vertex the same point D: but the py11.

ramid 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, HKLD is similar to the whole pyramid ABCD: and bek 41. 1. cause BF is equal to FC, the parallelogram EBFG is double k of

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

a triangle that is half of the parallelogram, these prisms are equal a 40. 11. a to one another; therefore the prism having the parallelogram

EBFG for its base, and the straight line KH opposite to it, is equal to the prism having the triangle GFC for its base, and the triangle HKL opposite to it; for they are of the same alti

tude, because they are between the parallel b planes ABC, HKL: B. XII. and it is manifest that each of these prisms is greater than either of the pyramids of which the triangles AEG, HKL are the b 15. 11. bases, and the vertices the points H, D; because, if EF be joined, the prism having the parallelogram EBFG for its base, and KH the straight line opposite to it, is greater than the pyramid of which the base is the triangle EBF, and vertex the point K; but this pyramid is equal to the pyramid the base of which is c C. 11 the triangle AEG, and vertex the point H; because they are contained by equal and similar planes: wherefore the prism having the parallelogram EBFG for its base, and opposite side KH, is greater than the pyramid of which the base is the triane, gle AEG, and vertex the point H: and the prism of which the base is the parallelogram EBFG, and opposite side KH, is equal to the.prism having the triangle GFC for its base, and HKL the triangle opposite to it: and the pyramid of which the base is the triangle AEG, and vertex H, is equal to the pyramid of which the base is the triangle HKL, and vertex D: therefore the two prisms before mentioned are greater than the two pyramids of which the bases are the triangles AEG, HKL, and vertices the points H, D. Therefore the whole pyramid of which the base is the triangle ABC, and vertex the point D, is divided into two equal pyramids similar to one another, and to the whole pyramid ; and into two equal prisms; and the two prisms are toge ther greater than half of the whole pyramid. Q. E. D. .

« ForrigeFortsett »