C. II. 5.

C

Book XII. is to the space S. therefore as the circle ABCD is to the space S, fo is the polygon AXBOCPDR to the polygon EKFLGMHN. but 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 impoffible. 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 fquare 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 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. but as the space + T is to the circle ABCD, fo is the circle EFGH to fome space, which must be lefs & than the circle ABCD, because the space T is greater, by hypothesis, than the circle EFGH. therefore as the fquare 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 fquares 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 fome of the following Propofitions.

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

E

VERY pyramid having a triangular base, may be see N, divided into two equal and fimilar pyramids having triangular bases, and which are fimilar to the whole pyramid; and into two equal prifms which together are greater than half of the whole pyramid.

D

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 fimilar pyramids having triangular bafes, and fimilar to the whole; and into two equal prifms 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 to DB. for the fame reason, HK is parallel to AB. therefore HEBK is a parallelogram, and HK equal to EB. but EB is equal to AE; therefore alfo AE is B equal to HK. and AH is equal to HD;

b

K

H

L

E

G

a. 2. 6.

b. 34. I.

F

wherefore EA, AH are equal to KH, HD, each to each; and the

C

angle EAH is equal to the angle KHD; therefore the base EH is c. 29. 1.

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

d. 4. I.

f. C. II.

f

Book XII. equal to the bafe KD, and the triangle AEH equal and fimilar to the triangle HKD. for the fame reason, the triangle AGH is equal and fimilar to the triangle HLD. and because the two ftraight lines EH, HG which meet one another are parallel to KD, DL that meet one another, and are not in the fame, plane with them, they e. 10. 11. contain equal 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 bafe EG is equal to the base KL, and the triangle EHG equal d and fimilar to the triangle KDL. for the fame reason, the triangle AEG is also equal and fimilar to the triangle HKL. Therefore the pyramid of which the base is the triangle AEG, and of which the vertex is the point H, is equal 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 fide of the triangle ADB, the triangle ADB is equiangular to the triangle HDK, and their fides are proportionals %, therefore the triangle ADB is fimilar to the triangle HDK. and for the fame reason, the triangle DBC is fimilar to the triangle DKL; and the triangle ADC to the triangle HDL; and alfo the triangle ABC to the triangle AEG. but the triangle AEG is fimilar to B the triangle HKL, as before was proved,

g. 4. 6.

h. 21.6.

11.Def.II.

K

E

D

H

L

G

F

therefore the triangle ABC is fimilar 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 fimilar to the pyramid of which the base is the triangle HKL, and vertex the fame point D. but the pyramid of which the bafe is the triangle HKL, and vertex the point D, is fimilar, 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 fimilar to the pyramid of which the bafe is the triangle AEG and vertex H. therefore each of the pyramids AEGH, HKLD is fimilar to the whole pyramid ABCD. and because BF is equal to FC, the parallelogram k. 41. I. EBFG is double of the triangle GFC. but when there are two prifms of the fame altitude, of which one has a parallelogram for its base, and the other a triangle that is half of the parallelogram,

m

1. 40. II.

these prisms are equal 1 to one another; therefore the prifm having Book XII. the parallelogram EBFG for its base, and the straight line KH oppofite to it, is equal to the prism having the triangle GFC for its bafe, and the triangle HKL oppofite to it; for they are of the fame altitude, because they are between the parallel planes ABC, m. 15. II. HKL. and it is manifeft that each of these prifms is greater than either of the pyramids of which the triangles AEG, HKL are the bafes, and the vertices the points Ĥ, D; because if EF be joined, the prism having the parallelogram EBFG for its base, and KH the ftraight line oppofite 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 the triangle AEG, f. C. 11. and vertex the point H; because they are contained by equal and fimilar planes. wherefore the prism having the parallelogram EBFG for its bafe, and oppofite fide KH, is greater than the pyramid of which the bafe is the triangle AEG, and vertex the point H. and the prism of which the base is the parallelogram EBFG, and oppofite fide KH is equal to the prism having the triangle GFC for its base, and HKL the triangle oppofite to it; and the pyramid of which the bafe 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 prifms before-mentioned are greater than the two pyramids of which the bafes 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 fimilar to one another, and to the whole pyramid; and into two equal prifms; and the two prisms are together greater than half of the whole pyramid. Q. E. D.

Book XII.

See N.

a. 2. 6.

IT

PROP. IV. THEOR.

F there be two pyramids of the same altitude, upon triangular bases, and each of them be divided into two equal pyramids fimilar to the whole pyramid, and alfo into two equal prifms; and if each of these pyramids be divided in the fame manner as the first two, and so on. as the base of one of the first two pyramids is to the base of the other, fo fhall all the prifms in one of them be to all the prifms in the other, that are produced by the fame number of divisions.

Let there be two pyramids of the fame altitude upon the triangular bases ABC, DEF, and having their vertices in the points G, H; and let each of them be divided into two equal pyramids fimilar to the whole, and into two equal prifms; and let each of the pyramids thus made be conceived to be divided in the like manner, and so on. as the base ABC is to the base DEF, so are all the prifms in the pyramid ABCG to all the prisms in the pyramid DEFH made by the fame number of divifions.

Make the fame construction as in the foregoing propofition. and becaufe BX is equal to XC, and AL to LC, therefore XL is parallel a to AB, and the triangle ABC fimilar to the triangle LXC. for the fame reason, the triangle DEF is fimilar to RVF. and because BC is double of CX, and EF double of FV, therefore BC is to CX, as EF to FV. and upon BC, CX are described the fimilar and fimilarly fituated rectilineal figures ABC, LXC; and upon EF, FV, in like manner, are described the similar figures DEF, RVF. therefore as the triangle ABC is to the triangle LXC, so is the triangle DEF to the triangle RVF, and, by permutation, as the triangle ABC to the triangle DEF, fo is the triangle LXC to the triangle RVF. and because the planes ABC, OMN, as alfo the planes c. 15. 11. DEF, STY are parallel ©, the perpendiculars drawn from the points

b. 22. 6.

G, H to the bafes ABC, DEF, which, by the Hypothesis, are equal d. 17. 11. to one another, fhall be cut each into two equal parts by the planes OMN, STY, because the ftraight lines GC, HF are cut into two equal parts in the points N, Y by the fame planes. therefore the prisms LXCOMN, RVFSTY are of the fame altitude; and