4. 6.!

fame Circumference; and the Angle FLG ist egual +21.3. to the Angie FNG: Therefore the Angle AMB is equal to the Angle FNG. But the Right Angle BA M is equal to the Right Angle G F N; where- ‡ 31.3. fore the other Angle fhall be equal to the other Angle: And fo the Triangle AMB is equiangular to the Triangle GFN; and confequently, as * B M is to * GN, fo is BA to G F. But the Proportion of the Square of B M to the Square of G N, is duplicate of the Proportion of BM to GN; and the Proportion of the Polygon A B C D E to the Polygon FGH KL, is + duplicate of the Proportion of BA to FG: Where- + 20. 6. fore, as the Square of B M is to the Square of G N, fo is the Polygon ABCDE to the Polygon FGHK L. Therefore, fimilar Polygons, infcribed in Circles, are to another as the Squares of the Diameters of the Circles; which was to be demonftrated.

元

K

If there be two unequal Mag-
nitudes propofed, and from
the greater be taken a Part
greater than its Half; and
if from what remains there
be again taken a Part
greater than half this Re- H
mainder ; and again from.
this laft Remainder a Part
greater than its Half; and
if this be done continually;
there will remain at last a
Magnitude that shall be less
than the leffer of the pro-
pofed Magnitudes.

V

LET AB and C be two unequal Magnitudes whereof AB is the greater: Ifay, if from A B be taken a greater Part than Half, and from the Part remaining there be again taken a Part greater than its

Half

Half, and this be done continually, there will remain a Magnitude at last, that shall be less than the Mag-nitude C.

For C being fome Number of Times multiplied, will become greater than the Magnitude A B. Let it be multiplied, and let D E be a Multiple of C greater than A B; divide D E into Parts D F, FG, GE, each equal to C, and take B H, a Part greater than half of A B, from A B, and again from A H, the Part HK greater than half A H, and from AK a Part greater than half A K, and so on, until the Divifions that are in A B, are equal in Number to the Divifions in DE: therefore let the Divifions A K, KH, HB, be equal in Number to the Divifions D F, F G, GE: Then because D E is greater than A B, and the Part EG taken from ED is lefs than half thereof, and the Part B H, greater than half of A B, is taken from it; the Part remaining, DG, fhall be greater than the Part remaining HA. Again, because G D is greater than HA; and G F being half of G D, is taken from the fame; and HK, being greater than half H A, is taken from this likewife; the Part remaining, FD, fhall be greater than the Part remaining, A K. But FD is equal to C; therefore C is greater than A K; and fo the Magnitude A K is leffer than C: Therefore, the Magnitude AK, being the Part remaining of the Magnitude A B, is less than the leffer propofed Magnitude C; which was to be demonftrated. If the Halves of the Magnitudes should have been taken, we demonftrate this after the fame manner. This is the firft Propofition of the tenth Book.

PROPOSITION II.

THEOREM.

Circles are to each other as the Squares of their Diameters.

LET ABD, EFGH, be Circles, whofe Dia meters are BD, FH. I fay, as, the Square of BD is to the Square of F H, fo the Circle ABCD is to the Circle EFCK.

For,

For, if it be not to, the Square of BD fhall be to the Square of F H, as the Circle ABCD is to fome Space either lefs or greater than the Circle E F G H. First, let it be to a Space S, lefs than the Circle EFGH; and let the Square E F G H be described in the Circle. This Square EFGH will be greater than half the Circle E F G H ; because, if we draw Tangents to the Circle thro' the Points E, F, G, H, the Square EFGH will be half that defcribed about the Circle: But the Circle is less than the Square described about it; therefore the Square EFGH is greater than half the Circle EFGH. Let the Circumferences E F, F G, GH, HE, be bifected in the Points, K L M N; and join EK, KF, FL, LG, G M, MH, HN, NE: Then each of the Triangles E K F, FLG, GM H, HNE, will be greater than one half of the Segement of the * 41. 1. Circle it ftands in; becaufe, if Tangents at the Circle be drawn thro' the Points K, L, M, N, and the Parallelograms that are on the Righr Lines E F, FG, GH, HE, be compleated, each of the Triangles E K F, FLG, GMH, HN E, is half of each of the correfponding Parallelograms: But the Segment is lefs than the Parallelogram; wherefore, each of the Triangles EK F, FLG, GMH, HNE, is greater than one half of the Segment of the Circle in which it ftands: Therefore, if thefe Circumferences be again bifected, and Right Lines be drawn joining the Points of Bifection, and you do thus continually, there will at laft remain Segments of the Circle that fhall be lefs than the Excefs, by which the Circle E F G H exceeds the Space S. For it is demonftrated, in the foregoing Lemma, that, if there be two unequal Magnitudes propofed, and if from the greater a Part greater than half be taken, and again from the Part remaining a Part greater than halt be taken, and you do thus continually, there will at laft remain a Magnitude that will be less than the leffer propofed Magnitude. Let the Segments of the Circle EFGH on the Right Lines EK, KF, FL, LG, GM, MH, HN, NE be those which are lefs than the Excefs, whereby the Circle E F G H exceeds the Space S; and then the remaining Polygon EKFLGMH N fhall be greater than the Space S. Also, defcribe the Polygon

R

.

AX

† 11. 5.

14. 5.

AXBOCPDR in the Circle A B CD, fimilar to the Polygon E KFLGMHN. Wherefore, as the Square of BD is to the Square of FH, fo is 1 of this, the Polygon A X BOCPDR to the * Polygon EKFLGMHN. But as the Square of B D is to the Square of FH, fo is the Circle A B C D to the From Hyp· Space S‡: Wherefore, as the Circle A B C D is to Space S, fo is + the Polygon A X BOCPDR to the Polygon EKFLGMHN. But the Circle A BCD is greater than the Polygon in it; wherefore the Space S fhall be * also greater than the Polygon From Hyp. EK FLGMHN: But it is lefs likewife; which is abfurd; therefore the Square of BD to the Square of F H, is not as the Circle A B C D to fome Space lefs than the Circle E F G H. After the fame manner we likewife demonftrate, that the Square of FH to he Square of B D, is not as the Circle E F G H to fome Space T less than the Circle ABCD. Laftly, I fay, the Square of BD to the Square of F G, is not as the Circle A B C D to fome Space greater than the Circle E F G H: For, if it be poffible, let it be fo, and let the Space S be greater than the Circle EFGH: Then it fhall be (by Inverfion) as the Square of FH is to the Square of BD, fo is the Space S to the Circle A B C D. But, because S is greater than the Circle E F G H, the Space S shall be to the Circle A B C D, as the Circle E F G H is to fome Space T lefs than the Circle A B CD: Therefore, as the Square of FH is to the Square of B D, fo is the Circle EF GH to fome Space T 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 A B C D to fome Space greater than the Circle E FGH: But it also has been proved, that the Square of B D to the Square of FH, is not as the Circle A B C D to fome Space lefs than the Circle E F G H: Wherefore, as the Square of BD is to the Square of F H, fo fhall the Circle ABCD be to the Circle EFGH. Wherefore, Circles are to each other as the Squares of their Diameters; which was to be demonstrated.

*

PRO

PROPOSITION III.
THEOREM.

Every Pyramid having a triangular Base, 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 Prisms, which two Prisms are greater
than the Half of the whole Pyramid.
LET there be a Pyramid, whofe Bafe is the Tri-

angle 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.

4. 1.

For, bifect A B, B ̊C, C A, a D, D B, D C, in the Points E, F, G, H, K, L; and join E H, EG, GH, HK, KL, LH, EK, KF, FG: Then, becaufe A E is equal to E B, and A H to HD; EH fhall be pa- • 2. 6% rallel to DB; for the fame Reafon HK alfo is parallel to A B; therefore H E BK is a Parallelogram: and fo H K is equal to EB: But E B is equal to AE; † 346 3. therefore A E fhall be alfo equal to HK; but A His equal to HD; wherefore the two Sides A E, AH, are equal to the two Sides K H, HD, each to each, and the Angle E A H is equal to the Angle K HD; † 29. 1. wherefore the Bafe EH is equal to the Bafe KD; and fo the Triangle A E H is equal and fimilar to the Triangle HK D. For the fame Reafon the Triangle AH G fhall also be equal and fimilar to the Triangle HDL; and because the two Right Lines E H, HG, touching each other, are parallel to the two Right Lines KD, DL, touching each other, and not in the fame Plane with them, they fhall contain † equal Angles. † 10. ats Therefore the Angle E H G is equal to the Angle KDL. Again, because the two Sides E H, H G, are equal to the two Sides K D, DL, each to each; and the Angle E HG is equal to the Angle KDL; the Bafe EG fhall be equal to the Bafe KL; and herefore the Triangle EH G is equal and fimilar to the Triangle KD L. For the fame Reason, the TriR 2

anglé

4. Is