*

Polygon EKFLGMHN. Wherefore as the Square of BD is to the Square of F H, fo is the Polygon AX BOCPDR to the Polygon EKFLGMHN. But of this as the Square of BD is to the Square of FH, fo is the Circle ABCD to the Space S. Wherefore as the Circle ABCD is to the Space S, fo is † the Polygon † 11. 5. AXBOCPDR to the Polygon EKFDGMÍN. But the Circle ABCD is greater than the Polygon in it. Wherefore the Space S fhall be + alfo greater than the Polygon EKFLGMHN, but it is lefst From the likewife; which is abfurd. Therefore the Square of Hyp BD to the Square of FH, is not as the Circle ABCD to fome Space lefs than the Circle EFGH. After the fame manner we likewife demonftrate that the Square of FH to the Square of BD is not as the Circle EFGH, to fome Space lefs than the Circle ABCD. Laftly, I fay, the Square of BD to the Square of FH is not as the Circle ABCD, to fome Space greater than the Circle EF GH; for if it be poffible, let it be fo, and let the Space S be greater than the Circle EF GH; then fhall it be (by Inverfions) as the Square of F H is to the Square of BD, fo is the Space S to the Circle ABCD. But because S is greater than the Circle E F G H, the Space fhall be to the Circle ABCD, as the Circle EFGH is to fome Space less than the Circle ABCD. Therefore, as the Square of FH is to the Square of BD, fo is the Circle * 11. 5. EF GH to fome Space lefs than the Circle ABCD; which has been demonftrated to be impoffible. Wherefore the Square of BD to the Square of FH, is not as the Circle ABCD to fome Space greater than the Circle EF GH. But this also has been proved, that the Square of BD to the Square of FH, is not as the Circle ABCD to fome Space lefs than the Circle EFGH. Wherefore as the Square of BD is to the Square of F H, fo fhall the Circle ABCD be to the Circle EF GH. Wherefore Circles are to each other as the Squares of their Diameters; which was to be demonftrated.

*

PRO.

2. 6.

† 34. 1.

29. I.

4. I.

PROPOSITION III.
THEOREM.

Every Pyramid having a triangular Base may be
divided into two Pyramids, equal and fimilar
to one another, having triangular Bases, and
fimilar to the whole Pyramid, and into two
equal Prifms, which two Prifms are greater
than the half of the whole Pyramid.

ET there be a Pyramid, whose Bafe is the Triangle ABC; and Vertex the Point D. I fay, the Pyramid ABCD may be divided into two Pyramids equal and fimilar to one another, having triangular Bafes, and fimilar to the whole; and into two equal Prifms; which two Prisms are greater than the half of the whole Pyramid.

For bifect 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 fhall be* parallel to DB. For the fame Reafon, HK alfo is parallel to AB. Therefore HEBK is a Parallelogram; and fo HK is † equal to EB, but EB is equal to AE. Therefore AE fhall be alfo 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 the Angle EAH is equal to the Angle KHD: Wherefore the Base EH is* equal to the Base KD: And fo the Triangle AEH is equal and fimilar to the Triangle HKD. For the fame Reason, the Triangle AHG fhall alfo 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 fame 10. 11. Plane with them, they fhall 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 Bafe EG fhall be equal to the Bafe KL: And therefore, the Triangle EHG is equal and fimilar to the Triangle KDL. For the fame Reason, the Trei

4. I,

angle

angle AEG is alfo equal and fimilar to the Triangle HKL. Wherefore the Pyramid whofe Bafe is the Triangle AEG, and Vertex the Point H, is equal and fimilar to the Pyramid whofe Base is the Triangle HKL, and Vertex the Point D. And because HK is drawn parallel to the Side AB of the Triangle ADB, the Triangle ADB fhall 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 fame Reason, the Triangle DBC is fimilar to the Triangle DKL; and the Triangle AHG to the Triangle DHL. And fince the two Right Lines B.A, AC, touching each other, are parallel to the two Lines KH, 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 AC, as KH is to HL. Wherefore the Triangle ABC is fimilar to the Triangle HKL; and fo the Pyramid, whose Base is the Triangle 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, whose Base is the Triangle HKL, and . Vertex the Point D, has been proved fimilar to the Pyramid whose Bafe is the Triangle AEG, and Vertex the Point H. Therefore the Pyramid whofe 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 fince 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; thofe Prisms will be equal to one another. Therefore the Prifm con-† 40. 11. tained under the two Triangles BKF, EHG, and the three Parallelograms EBFG, EBKH, KHGF, is equal to the Priẩm contained under the two Triangles GFC, HKL, and the three Parallelograms KFCL, LCGH, HKFG. And it is manifeft that each of those Prifms, the Base of one of which is the Parallelogram EBGF, and oppofite Base to that the Right Line

R

KH,

KH, and the Bafe of the other the Triangle GFC and the oppofite Base to this, the Triangle KLH, are greater than either of the Pyramids, whofe Bafes are the Triangles AEG, HKL, and Vertices the Points H and D. For fince, if the Right Lines EF, EH be joined, the Prism, whose Base is the Parallelogram EBFG, and oppofite Base to that the Right Line KH, is greater than the Pyramid, whofe Bafe is the Triangle EBF, and Vertex the Point K. But the Pyramid whofe Bafe is the Triangle E B F, 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 fimilar Planes. Wherefore the Prifm whofe Bafe is the Parallelogram EBFG, and opposite Base to it the Right Line HK, is greater than the Pyramid whofe Bafe is the Triangle AEG, and Vertex the Point H. But the Prism whose Base is the Parallelogram EBF G, and oppofite Bafe 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 Prifms aforefaid, are greater than the faid two Pyramids, whofe Bafes are the Triangles AEG, HKL, and Vertices the Points H, D. And fo the whole Pyramid whofe Bafe 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 Prifms; which two Prifms together are greater than half of the whole Pyramid. Therefore, Every Pyramid having a triangular Bafe may be divided into two Pyramids, equal and fimilar to one another, having triangular Bafes, and fimilar to the whole Pyramid, and into two equal Prifms, which two Prifms are greater than the half of the whole Pyramid; which was to be demonftrated.

PRO

PROPOSITION IV.

THEOREM.

If there are two Pyramids of the fame Altitude, baving triangular Bafes, and each of them be divided into two Pyramids, equal to one another, and fimilar to the whole, as also into two equal Prisms; and if in like manner each of the two Pyramids, made by the former Divifion, be divided, and this be done continually; then as the Bafe of one Pyramid is to the Bafe of the other Pyramid, fo are all the Prifms that are in one Pyramid to all the Prifms that are in the other Pyramid being equal in Multitude.

ET there be two Pyramids of the fame Altitude,

having the triangular Bafes ABC, DEF, whose Vertices are the Points G, H, and let each of them be divided into two Pyramids, equal to one another, and fimilar to the whole, and into two equal Prifms; and if in like manner each of the Pyramids made by the former Divifion be conceived to be divided, and this be done continually. I fay, as the Bafe ABC is to the Bafe DEF, fo are all the Prisms that are in the Pyramid ABCG to all the Prisms that are in the Pyramid DEF H, being equal in Multitude.

For fince B X is equal to X C, and AL to LC, XL fhall be parallel to AB, and the Triangle ABC* 2. 6, fimilar to the Triangle LXC. For the fame Reason the Triangle DEF fhall be alfo fimilar to the Triangle RQF. And because BC is double to CX, and EF to FQ, it fhall be as BC is to CX, fo is EF to FQ. And fince there are defcribed upon BC, CX, Right-lined Figures ABC, LXC, fimilar and alike fituate, and upon EF, FQ, Right-lined Figures DEF, RQF, fimilar and alike fituate. Therefore as the Triangle BAC is to the Triangle LXC, fo is + the † 22.6. Triangle DEF to the Triangle RQF; and (by Alternation) as the Triangle ABC is to the Triangle DEF, fo is the Triangle LX C to the Triangle ROF. But as the Triangle LXC is to the Triangle RQF,

fo is the Prifm, whose Base is the Triangle LX C, ‡ 28. and and oppofite Base to that the Triangle OMN, to the 32. 11.

R 2

Prism,