PROP. XL. THEOR.

Itude, the bale trong of which is a parallelogram, [F there be two triangular prifms of the fame alti

and the base of the other a triangle; if the parallelogram be double of the triangle, the prisms fhall be equal to one another.

Let the prifms ABCDEF, GHKLMN be of the fame altitude, the first whereof is contained by the two triangles ABE, CDF, and the three parallelograms AD, DE, EC; and the other by the two triangles GHK, LMN and the three parallelograms LH, HN, NG; and let one of them have a parallelogram AF, and the other a triangle GHK for its base; if the parallelogram AF be double of the triangle GHK, the prism ABCDEF is equal to the prism GHKLMN.

Complete the folids AX, GO; and because the parallelogram AF is double of the triangle GHK; and the parallelo

Book XI.

gram HK double a of the fame triangle; therefore the paral- a 34. I. lelogram AF is equal to HK. But folid parallelepipeds upon equal bases, and of the fame altitude, are equal b to one an- b 31. II. other. Therefore the folid AX is equal to the folid GO; and the prifm ABCDEF is half of the folid AX; and the prism GHKLMN half of the folid GO. Therefore the prism ABCDEF is equal to the prism GHKLMN. Wherefore, if there be two, &c. Q. E. D.

c 28. Ir.

THE

LEMMA I.

Which is the first propofition of the tenth book, and is neceffary to fome of the propofitions of this book.

I'

F from the greater of two unequal magnitudes, there be taken more than its half, and from the remainder more than its half; and fo on: There fhall at length remain a magnitude less than the leaft of the proposed magnitudes.

Let AB and C be two unequal magnitudes, of which AB is
the greater. If from AB there be taken more
than its half, and from the remainder more A
than its half, and fo on; there fhall at length
remain a magnitude less than C.

For C may be multiplied fo as at length to
become
than AB.
greater
Let it be fo multi-

D

K

F

H

G

· plied, and let DE its multiple be greater than
AB, and let DE be divided into DF, FG, GE,
each equal to C. From AB take BH greater
than its half, and from the remainder AH
take HK greater than its half, and fo on, until
there be as many divifions in AB as there are
in DE: And let the divifions in AB be AK,
KH, HB; and the divifions in ED be DF, FG B CE
GE. And because DE is greater than AB, and

that

that EG taken from DE is not greater than its half, but BH Book XII. taken from AB is greater than its half; therefore the remainder GD is greater than the remainder HA. Again, because GD is greater than HA, and that GF is not greater than the half of GD, but HK is greater than the half of HA; therefore the remainder FD is greater than the remainder AK. And FD is equal to C, therefore C is greater than AK; that is, AK is less than C. Q. E.D.

And if only the halves be taken away, the fame thing may in the fame way be demonftrated.

SIMILAR

PROP. I. THEOR.

IMILAR polygons infcribed in circles, are to one
another as the squares of their diameters.

Let ABCDE, FGHKL be two circles, and in them the fimilar polygons ABCDE, FGHKL; and let BM, GN be the diameters of the circles: As the square of BM is to the fquare of GN, fo is the polygon ABCDE to the polygon FGHKL.

Join BE, AM, GL, FN: And because the polygon ABCDE is fimilar to the polygon FGHKL, and fimilar polygons are divided into fimilar triangles; the triangles ABE, FGL,are fimilar

C

and equiangular b; and therefore the angle AEB is equal to the 6 6.6. angle FLG: But AEB is equal to AMB, because they stand up- c 21, 3. on the same circumference; and the angle FLG is for the fame reason, equal to the angle FNG: Therefore alfo the angle AMB is equal to FNG: And the right angle BAM is equal to the right dangle GFN; wherefore the remaining angles in the tri- d 31, 3. angles ABM, FGN are equal, and they are equiangular to one

2

another :

e 4. 6.

Book XII. another: Therefore as BM to GN, fo e is BA to GF; and therefore the duplicate ratio of BM to GN, is the fame f with the duplicate ratio of BA to GF: But the ratio of the fquare of BM to the fquare of GN, is the duplicate & ratio of that which BM has to GÑ; and the ratio of the polygon ABCDE to the polygon

f 10. def. 5. & 22. 5.

20.6.

A

See N.

B

FGHKL is the duplicate g of that which BA has to GF: Therefore as the fquare of BM to the fquare of GN, fo is the polygon ABCDE to the polygon FGHKL. Wherefore fimilar polygons, &c. Q. E. D.

PROP. II. THEOR.

IRCLES are to one another as the fquares of their diameters.

C1

Let ABCD, EFGH be two circles, and BD, FH their diameters: As the fquare of BD to the fquare of FH, so is the circle ABCD, to the circle EFGH.

For, if it be not so, the fquare of BD fhall be to the square of FH, as the circle ABCD is to fome space either lefs than the circle EFGH, or greater than it *. First let it be to a space S less than the circle EFGH; and in the circle EFGH defcribe the fquare EFGH: This fquare is greater than half of the circle EFGH; because if, through the points E, F, G, H, there be drawn tangents to the circle, the fquare

*For there is fome fquare equal to the circle ABCD; let P be the fide of it, and to three straight lines BD, FH and P, there can be a fourth propor tional; let this be Q: Therefore the fquares of thefe four straight lines are

I

proportionals; that is, to the fquares of BD, FH and the circle ABCD, it is poffible there may be a fourth proportional. Let this be S. And in like manner are to be understood fome things in fome of the following propofitions.

fquare EFGH is half a of the square described about the circle; Book XII.
and the circle is lefs than the fquare described about it ; there- a 41.
fore the fquare EFGH is greater than half of the circle. Di-
vide the circumferences EF, FG, GH, HE, each into two equal
parts in the points K, L, M, N, and join EK, KF, FL, LG, GM,
MH, HN, NE: Therefore each of the triangles EKF, FLG,
GMH, HNE is greater than half of the fegment of the circle.
it stands in; because, if straight lines touching the circle be
drawn through the points K, L, M, N, and parallelograms up-
on the ftraight lines EF, FG, GH, HE, be completed; each
of the triangles EKF, FLG, GMH, HNE fhall be the half a a 41.
of the parallelogram in which it is: But every fegment is lefs
than the parallelogram in which it is: Wherefore each of the
triangles EKF, FLG, GMH, HNE is greater than half the
fegment of the circle which contains it: And if thefe cir-
cumferences before named be divided each into two equal parts,
and their extremities be joined by ftraight lines, by continuing

to do this, there will at length remain fegments of the circle
which, together, fhall be less than the excefs of the circle EFGH
above the space S: Becaufe, by the preceding Lemma, if
from the greater of two unequal magnitudes there be taken
more than its half, and from the remainder more than its
half, and so on, there fhall at length remain a magnitude lefs
than the least of the propofed magnitudes. Let then the feg-
ments EK, KF, FL, LG, GM, MH, HN, NE be thofe that
remain and are together less than the excess of the circle EFGH
above S: Therefore the reft of the circle, viz. the polygon
EKFLGMHN, is greater than the space S. Defcribe likewife
in the circle ABCD the polygon AXBOCPDR fimilar to the
polygon EKFLGMHN: As therefore, the square of BD is to
the fquare of FH, fob is the polygon AXBOCPDR to the 1. 12.
polygon EKFLGMHN: But the fquare of BD is also to the