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B.

Therefore

For bifect AB, BC, CA, AD, DB, DC in the points F, F, G, H, K, L, and join E H, EG, GH, HK, KL, LH, EK, KF, FG. Then because AE is equal to E E, and AH to HD; [by 2.6.] EH will be parallel to pв. By the fame realon HK is alfo parallel to A B: HEBK is a parallelogram; and fo [by 34. 1.] HK is equal to E B. But EB is equal to AE; and therefore A E will be equal to HK; but AH is equal to HD: Therefore the two fides AE, AH are equal to the two fides KH, HD, each to each, and [by 29. 1.] the angle EA H is equal to the angle KHD. Wherefore the bafe EH [by 4.1.] is equal to the base KD; and fo the triangle AEH is equal and fimilar to the triangle HKD. By the fame reafon the triangle AHG is equal and fimilar to the triangle HLD. And because the two right lines EH, HG B touching one another, are

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parallel to two right lines KD, DL touching one another, and not in the fame plane; they will contain equal angles [by 10. 11.] Therefore the angle EHG is equal to the angle KDL. Again, because the two fides EH, GH are equal to the two fides KD, DL, each to each, and the angle E HG is equal to the angle KDL; the base EG will be equal to the bafe KL; therefore the triangle EHG is equal and fimilar to the triangle KL. By the fame reafon the triangle AEG is equal and similar to the triangle HKL. Wherefore the pyramid whofe base is the triangle AEG, and vertex the point H, is equal and fimilar to the pyramid whose base is the triangle H KL, and vertex the point D. And because HK is drawn parallel to one fide A B of the triangle ADB; [by 29. 1.] the triangle ADB is equiangular to the triangle DH K; and [by 4. 6.] they have proportional fides: Therefore the triangle ADB is fimilar to the triangle DH K. And by the fame reafon the triangle DBC is fimilar to the triangle DKL, and the triangle ADC to the triangle DHL. And fince two right lines BA, A C touching one another, are parallel to two right lines KH, HL touching one another, not be

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Book XII. ing in the fame plane: [by 10. 11.] they will contain equal angles: Therefore the angle BAC is equal to the an gle KHL: And it is as BA to AC, fo is KH to HL: Therefore [by 6. 6.] the triangle ABC is fimilar to the triangle H KL; and fo the pyramid, whose base is the tri angle ABC, and vertex the point D, is fimilar to the pyramid, whofe bafe is the triangle HKL, and vertex the point D. But the pyramid, whofe bafe is the triangle HKL, and vertex the point D, has been proved to be equal to the pyramid whose base is the triangle AEG, and ver tex the point : And therefore the pyramid, whose base is the triangle ABC and vertex the point D, is fimilar to the pyramid, whofe bafe is the triangle A EG, and vertex the point H. Therefore each of the pyramids AEGH, HELD is fimilar to the whole pyramid ABCD. And because BF is equal to FC; the parallelogram EBFG will be [by 41. 1.] double to the triangle GFC. And because if there be two prifms of the fame altitude, one of which has a parallelogram for its base, and the other a triangle, and the parallelogram be double to the triangle; thefe prifms [by 40. 1.] are equal to one another: Therefore the prifms contained under the two triangles BK F, E H G, and the three parallelograms E BF G E BK H, KHGF are equal to the prifm contained under the two triangles GFC, HKL, and the three parallelograms KFCL, LCGH, HKFG. And it is manifeft that each of the prifms, that whofe bafe is the parallelogram EBF G, and oppofite to it the right line HK, and that whofe bafe is the triangle GFC, and oppofite bafe the tangle KLH, is greater than either of the pyramids whofe bafes are the triangles AEG, HKL, and vertexes the points H, p. Because if the right lines IF, EK be joined, the prifm, whose base is the parallelo gram EBFG, and oppofite to it the right line HK, is greater than the pyramid whofe bafe is the triangle EBF, and vertex the point K. But the pyramid, whose base is the triangle EBF, and vertex the point K, [by 10. def. 11.] is equal to the pyramid, whofe bafe is the triangle AEG, and vertex the point H; for they are contained under equal and fimilar planes. Therefore the prifm, whose bafe is the parallelogram EBFG, and the right line HK oppofite to it, is greater than the pyramid, whose base is the triangle AEG, and vertex the point : But the prifm whofe

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whose base is the parallelogram E B FG, and the right line HK is oppofite to it, is equal to the prifm whofe bafe is the triangle GFC, and the triangle HKL is oppofite to it; and the pyramid, whose base is the triangle AEG, and vertex the point H, is equal to the pyramid. whose base is the triangle HKL, and vertex the point D. Therefore the two prifms first mentioned, are greater than the two prisms whose bases are the triangles A E G, HK L, and vertexes the points H, D. Therefore the whole pyramid, whofe base is the triangle ABC, and vertex the point D, is divided into two equal pyramids fimilar to one another and to the whole, and into two equal prifms, which are greater than the whole pyramid. Which was to be demonftrated.

PROP. IV. THEOR.

If there be two prifms of the fame altitude, having triangular bafes, and either of them be divided into two pyramids equal to one another, and fimilar t the whole, and into two equal prifms; and if in like manner each of the pyramids made by the former divifion be divided, and this be continually done: then as the base of one of the pyramids is to the bafe of the other, fo are all the prifms that are in one of the pyramids, to all the prifms in the other pyramid being equal in number.

Let there be two pyramids of the fame altitude, having the triangular bases ABC, DEF, and vertexes 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 made by such a divifion be divided after the fame manner; and if this be always done: I fay as the base ABC is to the base DEF, so are all the prifms in the pyramid ABC G, to all the prifms that are in the pyramid DEFH, being equal in number.

For because Bx is equal to xc, and AL to LC; [by 2.6.] XL will be parallel to A B, and [by 4. 6,] the triangle ABC is fimilar to the triangle LXC. By the fame reafon the triangle DEF is fimilar to the triangle R QF; and because BC is double to cx, and EF is double to FQ, as BC is to cx, fo will EF be to F Q. And there are Bb4

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defcribed upon BC, CX the fimilar right lined figures ABC, LXC alike fituate, and upon EF, FQ the fimilar

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DEF, ROF:

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the triangle DEF to the triangle RQF, and inversely as the triangle ABC is to the triangle DEF, fo is the triangle LXC to the triangle R QF. But [as it is demonftrated below] as the triangle LXC is to the triangle R QF, fo is the prifm whose base is the triangle LXC, and the oppofite to it OMN, to the prifm whose base is the triangle R QF, and oppofite bafe sTv: Therefore [by 11. 5.] as the triangle ABC, is to the triangle DEF, fo is the prism whofe bafe is the triangle LXc and oppofite bafe o MN, to the prifm whofe bafe is the triangle RQF, and oppofite bafe ST V. And becaufe the two prifms in the pyramid ABCG are equal to one another, as alfo the two prisms in the pyramid DEFH; the prifm whofe bafe is the parallelogram KLX B, and the right line M o oppofite to it, will be to the prifm whose base is the triangle LXC, and oppofite bafe OMN, as the prifm whose base is the parallelogram EPR Q, and oppofite to it the right line s v, to the prifm whole bafe is the triangle RQF, and oppofite bafe is ST V. Wherefore by compofition [by 18. 5.] as the prifms KB LX MO, LXC MN O, are to the prism LXCMNO,, fo are the prifms PEQRST, ROFST V to the prifm ROESTV: And by alternation, as the prifms K BX LOM, LXCOM N are to the prifms PEQRST, RQFSTV, fo is the prifm LXC MNO, to the: prifm RQFSTY. But, it has been proved that as the prifm LXCM NO is to the prifm RQF STV, fo is the base LXC to the base RQF, and the base A B C to the base DEF: Therefore as the triangle ABC is to the triangle DEF, fo are the two prifms that are in the pyramid ABCG, to the two prifms in the pyramid DEFH. So likewise if the pyramids made by the divifion aforefaid be divided after the fame manner as O MNG, STVн, it will be as the bafe

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bafe O M N is to the base s T v, fo are the two prisms in the pyramid OM NG to the two prifms that are in the pyramid STVH. But as the bafe OMN is to the bafe s T v, so is the base A B C to the base DEF. And therefore as the base ABC is to the bafe DEF, fo are the two prifms in the pyramid ABCG to the two prisms that are in the pyramid DEFH, and the two prifms that are in the pyramid OMNG to the two prifms in the pyramid s TVH, and fo are four to four. And the fame thing is also proved in the other prifms made by the divifions of the pyramids A KLO, and DPR S, and of all the reft equal in number. Which was to be demonftrated.

LEMMA.

But we thus demonftrate that the triangle LXC is to the triangle ROF, as the prifm whose bafe is the triangle LXC, and oppofite bafe o M N, to the prifm whofe bafe is the triangle ROF, and oppofite bafe the triangle s T v.

For in the fame figure conceive perpendiculars to be drawn from the points G, H, to the planes of the triangles 'A BC, DE F, which will be equal to one another; because the pyramids themselves are supposed to have the fame altitude, and because the two right lines, viz. GC, and the perpendicular drawn from the point G, are cut by the parallel planes ABC, OMN; [by 17. 11.] they will be cut in the fame ratio. But GC is bifected by the plane OM N in the point N: Therefore the perpendicular drawn from the point G to the plane ABC will be bifected by the plane OмN. By the fame reafon the perpendicular drawn from the point H to the plane DEF, will be bifected by the plane sTV. But the perpendiculars drawn from the points G, H to the planes A B C, D E F are equal: And therefore the perpendiculars drawn from the triangles o MN, STV, to ABC, DEF, are equal: Wherefore the prifms whofe bafes are the triangles LXC, RQF, and oppofite bafes the, triangles OMN, STV, have the fame altitude: Wherefore [by 32. 11.] the folid parallelepipedons which are defcribed upon the faid prifms of equal altitudes, are to one another as their bafes: and fo are their halves: Therefore as the bafe LXC is to the bafe RQF, fo will the faid prifms be to one another. Which was to be demonstrated.

PROP.

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