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PROPOSITION III.

THEOR İM.
Every Pyramid having a triangular Base may be

divided into two Pyramids, equal and similar
to one another, having triangular Baless and
similar to the whole Pyramid, and into two
equal Prisms, which two Prisms are greater
than the half of the whole Pyramid:
ET there be a Pyramid, whose Base is the Tri-

angle ABC; and Vertex the Point D. I fayz the Pyramid ABCD may be divided into two Pyrainids equal and similar to one another, having triangular Bases, and fiinilar to the whole; and into two equal Prisms; which two Prisms are greater than the half of the whole Pyramid.

For bisect AB, BC, CA, AD, DB, DC, in the Points E, F, G, H, K, L, and join EH, EG, GH, HK, KL, LH, EK, KF, FG. Then because AE is equal to EB, and AH to HD, EH shall be * parallel to DB. For the same Reason, HK also is pa

rallel to AB. Therefore HEBK is a Parallelogram; * 34. 1. and fo HK ist equal to EB, but EB is equal to AE.

Therefore AE shall be also equal to HK, but AH is equal to HD. Wherefore the two Sides AE, AH

are equal to the two Sides KH, HD, each to each, and I 29. 1. the Angle EAH is equal to the Angle K HD:

Wherefore the Base EH is * equal to the Base KD:
And so the Triangle AEH is equal and fimilar to the
Triangle HKD. For the fame Reason, the Triangle
AHG shall also be equal and fimilar to the Triangle
HLD. And because the two Right Lines EH, HG,
touching each other, are parallel to the two Right Lines

KD, DL, touching each other, and not in the same # 10. 11. Plane with them, they shall contain + equal Angles.

Therefore the Angle EHG is equal to the Angle KDL. Again, because the two Sides EH, HG are equal to the two Sides KD, DL, each to each, and

the Angle EHG is equal to the Angle KDL, the 4. I, Base EG shall be * equal to the Base KL: And

therefore, the Triangle ÈHG is equal and similar to the Triangle KDL. For the same Reason, the Trei

angle

2. 6.

4 1.

angle AEG is also equal and fimilar to the Triangle HKL. Wherefore the Pyramid whose Base is the Triangle AEG, and Vertex the Point H, is equal and fimilar to the Pyramid whose Base is the Triangle HKL, and Vertex the Point D. And because HK is drawn parallel to the Side A B of the Triangle ADB, the Triangle ADB Thall be equiangular to the Triangle DKH, and they have their Sides proportional. Therefore the Triangle ADB is fimilar to the Triangle DHK. And for the same Reason, the Triangle ĎBC is similar to the Triangle DKL; and the Triangle AHG to the Triangle DHL. And since the two Right Lines B.A, AC, touching each other, are parallel to the two Lines K H, HL, touching each other, not being in the fame Plane with them, these shall contain equal Angles. Therefore the Angle BAC is equal to the Angle KHL. And BA is to A C, as KH is to HL. Wherefore the Triangle ABC is similar to the Triangle HKL; and so the Pyramid, whose Base is the Triangle ABC, and Vertex the Point D, is similar to the Pyramid, whose Base is the Triangle HKL, and Vertex the Point D. But the Pyramid, whose Base is the Triangle HKL, and Vertex the Point D, has been proved fimilar to the Pyramid whose Base is the Triangle AEG, and Vertex the Point H. Therefore the Pyramid whose Base is the Triangle ABC, and Vertex the Point D, is fimilar to the Pyramid whose Base is the Triangle AEG. and Vertex the Point H.' Wherefore both the Pyramids AEGH, HKLD, are fimilar to the whole Pyramid ABCD. And because BF is equal to FC, the Parallelogram EBFG will be double to the Triangle GFC. And since there are two Prisms of equal Altitude, one of which has that Parallelogram for a Base, and the other the Triangle, and the Parallelogram is double to the Triangle; those Prisms will be † equal to one another. Therefore the Prism con- + 40. 11. tained under the two Triangles B KF, EHG, and the three Parallelograms EBFG, EBKH, KHGF, is equal to the Prism contained under the two Triangles GFC, HKL, and the three Parallelograms KFCL, LCGH, HKFG. And it is manifest that each of those Prisms, the Base of one of which is the Parallelogram EBGF, and opposite Bafe to that the Right Line

R

ΚΗ,

KH, and the Base of the other the Triangle GFC and the opposite Base to this, the Triangle KLH, are greater than either of the Pyramids, whose Bases are the Triangles AEG, HKL, and Vertices the Points H and D. For since, if the Right Lines EF, EH be joined, the Prism, whose Base is the Parallelogram EBFG, and opposite Base to that the Right Line KH, is greater than the Pyramid, whose Base is the Triangle E BF, and Vertex the Point K. But the Pyramid whose Base is the Triangle E BF, and Vertex the Point K, is equal to the Pyramid whose Base is the Triangle AEG, and Vertex the Point H. For they are contained under equal and similar Planes. Wherefore the Prisın whose Base is the Parallelogram EBFG, and opposite Base to it the Right Line HK, is greater than the Pyramid whose Base is the Triangle AEG, and Vertex the Point H. But the Prism whose Base is the Parallelogram EBFG, and opposite Base to it the Right Line HK, is equal to the Prism whose Base is the Triangle GFC, and oppofite Bafe to this the Triangle HKL: 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 Prisms aforesaid, are greater than the said two Pyramids, whose Bases are the Triangles AEG, HKL, and Vertices the Points H, D. And fo the whole Pyramid whose Base is the Triangle ABC, and Vertex the Point D, is divided into two equal Pyramids, fimilar to each other and to the Whole : And into two equal Prisms; which two Prisms together are greater than half of the whole Pyramid. Therefore, Every Pyramid having a triangular Base may be divided into two Pyramids, equal and similar to one another, having triangular Bases, and similar to the whole Pyramid, and into two equal Prisms, which two Prisms are greater than the half of the whole Pyramid; which was to be demonstrated.

PRO

PROPOSITION IV.

THEORE M.
If there are two Pyramids of the same Altitude, bav.

ing triangular Bases, and each of them be divided
into two Pyramids, equal to one another, and fi-
milar to the whole, as also into two equal Prisms ;
and if in like manner each of the two Pyramids,
made by the former Division, be divided, and this
be done continually; then as the Base of one Pyra-
mid is to the Bafé' of the other Pyramid, so are all
the Prisms that are in one Pyramid to all the
Prisms that are in the other Pyramid being
equal in Multitude.

,
having the triangular Bases ABC, DEF, whose
Vertices are the Points G, H, and let each of them be
divided into two Pyramids, equal to one another, and
similar to the whole, and into two equal Prisms; and
if in like manner each of the Pyramids made by the
former Division be conceived to be divided, and this
be done continually. I say, as the Base ABC is to
the Base DEF, so are all the Prisms that are in the
Pyramid ABCG to all the Prisms that are in the
Pyramid DEFH, being equal in Multitude.

For since B X is equal to X C, and AL to LC, XL shall be * parallel to AB, and the Triangle ABC* 2. 6, fimilar to the Triangle LXC. For the fame Reason the Triangle DEF İhall be also similar to the Triangle ROF. And because BC is double to CX, and EF to F Q, it shall be as B C is to CX, so is É F to FO. And since there are described upon BC, CX, Right-lined Figures ABC, LXC, fimilar and alike situate, and upon EF, FQ, Right-lined Figures DEF, RQF, similar and alike fituate. Therefore as the Triangle BAC is to the Triangle LXC, so is the † 22.6. Triangle DEF to the Triangle RQF; and (by Alternation) as the Triangle ABC is to the Triangle DEF, so is the Triangle LXC to the Triangle ROF. But as the Triangle LXC is to the Triangle ROF, so is the Prism, whose Base is the Triangle LXC, † 23. and and opposite Base to that the Triangle OMN, to the 32. II.

Prism,

1

R2

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Prism whose Base is the Triangle RQF; and

oppofite Base to that the Triangle STY. Therefore as the Triangle A B C is to the Triangle DEF, so is * the Prism whose Base is the Triangle LX.C, and opposite Base to that the Triangle OMN, to the Prism whose Base is the Triangle RCF, and opposite Base to that the Triangle STY; and because the two Prisms that are in the Pyramid ABCG are equal to one another, as also those two that are in the Pyramid DEFH;it fhall be as the Prism whose Base is the Parallelogram KLX B, and opposite Bafe to that the Right Line MO, is to the Prism whofe Base is the Triangle LXC; and opposite Base to that the Triangle OMN, fo is the Prism whose Base is the Parallelogram EPRQ; and opposite Base to that the Right Line ST, to the Prism whose Base is the Triangle R QF, and opposite Bafe to that the Triangle STY. Therefore ( by compounding ) as the Prisms KBXLMO, LXCMNO, to the Prism LXCMNO, so the Prifms PEQRST, RQFSTY, to the Prism RQFSTY. And (by Alternation) as the Prisms KBXLMO, LXCMNO, to the Prisms PEQRST, RQFSTY, so the Prism LXCMNO, to the Prism RQFSTY; but as the Prism LXCMNO is to the Prism RQFSTY, fo has the Base LXC been proved to be to the Base RFQ; and so the Base ABC to the Base DEF. Therefore also as the Triangle ABC is to the Triangle DEF, so are the two Prisms that are in the Pyramid ABCG, to the two Prisms that are in the Pyramid DEFH. If in the fame Manner each of the Pyramids OMNG, STYH, made by the former Division, be divided, it shall be as the Base OMN is to the Base STY, so the two Prisms that are in the Pyramid O MNG, to the two Prisms that are in the Pyramid STYH. But as the Base OMN is to the Base STY, fo is the Base ABC to the Base DEF. Therefore as the Base ABC is to the Base DEF, so is the two Prisms that are in the Pyramid ABCG, to the two. Prisms that are in the Pyramid DEFH; and so the two Prisms that are in the Pyramid O MNG, to the two Prisms that are in the Pyramid STYH, and so the four to the four. We demonstrate the same of Prisms made by the Division of the Pyramids AKLO, DPRS,

and

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