#41. 1.

have the other Sides of one equal to the other Sides
of the other; and fo DT is equal to TG, and YT
to TS. Wherefore, if the Sides of the oppofite Planes
of a folid Parallelepipedon be divided into two equal
Parts, and Planes be drawn thro' their Sections; the
common Section of the Planes, and the Diameter of the
folid Parallelepipedon fhall divide each other into two
equal Parts; which was to be demonftrated.

PROPOSITION XL.
THEORE M.

If of two triangular Prifms, one ftanding on a Bafe,
which is a Parallelogram, and the other on a
Triangle, if their Altitudes from thefe Bafes are
equal, and the Parallelogram double to the Tri-
angle; then thofe Prifms are equal to each other.

LET ABCDEF, GHKLMN be two Prisms

of equal Altitude. The Bafe of one of which is the Parallelogram AF, and that of the other, the Triangle GHK, and let the Parallelogram AF be double to the Triangle GHK. I fay, the Prism ABCDEF is equal to the Prism GHKLMN.

For complete the Solids AX, GO. Then because the Parallelogram AF is double to the Triangle GHK, and fince the Parallelogram HK is * double to the Triangle GHK, the Parallelogram AF fhall be equal to the Parallelogram HK. But folid Parallelepipedons, that ftand upon equal Bafes, and 31 of this have the fame Altitude, are + equal to one another. Therefore the Solid A X is equal to the Solid GO. But the Prism ABCDEF is half the Solid AX, and 28 of this. the Prism GHKLMN ist half the Solid GO. Therefore the Prifm ABCDEF is equal to the Prism GHKLMN. Wherefore, if there be two Prisms having equal Altitudes, the Bafe of one of which is a Parallelogram, and that of the other a Triangle, and if the Parallelogram be double to the Triangle, the faid Prifms fhall be equal to each other,

The END of the ELEVENTH BOOK.

EUCLID's

EUCLID's

ELEMENTS.

BOOK XII.

237

PROPOSITION I.

THEOREM.

Similar Polygons, infcribed in Circles, are to one another as the Squares of the Diameters of the Circles.

L

ET ABCDE, FGHKL, be Circles, wherein are infcribed the fimilar Polygons ABCDE, FGH KL, and let BM, GN, be Diameters of the Circles. I fay, as the Square of B M is to the Square of GN, fo is the Polygon ABCDE to the Polygon F GHKL.

For join BE, AM, GL, FN. Then because the Polygon ABCDE is fimilar to the Polygon FGHKL, the Angle BAE is equal to the Angle GFL; and BA is to AE, as GF is to FL. Therefore the two Triangles BAE, GFL, have one Angle of the one equal to one Angle of the other, viz. the Angle BAE equal to the Angle GFL, and the Sides about the equal Angles proportional. Wherefore the Triangle ABE is equiangular to the Triangle FGL; * 6. 6.

and

† 21.3.

$31. 3.

4.6.

¥ 20.6.

and fo the Angle AEB is equal to the Angle FLG; But the Angle AEB is † equal to the Angle AMB; for they stand on the fame Circumference; and the Angle FLG is + equal to the Angle FNG. Therefore the Angle AMB is equal to the Angle FNG. But the Right Angle BAM is equal to the Right Angle GFN. Wherefore the other Angle fhall be equal to the other Angle. And fo the Triangle AMB is equiangular to the Triangle FGN; and confequently as BM is to GN, fo is BA to GF. But the Proportion of the Square of BM to the Square of GN, is duplicate of the Proportion of BM to GN; and the Proportion of the Polygon ABCDE to the Polygon F GHKL, ist duplicate of the Proportion of BA to GF. Wherefore, as the Square of BM is to the Square of GN, fo is the Polygon ABCDE to the Polygon F GHK L. Therefore, fimilar Polygons, infcribed in Circles, are to one another as the Squares of the Diameters of the Circles; which was to be demonftrated.

*

LE M M A.

If there be two unequal Mag-
nitudes propos'd, 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-
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
lefs than the leffer of the
propos'd Magnitudes.

ET AB and C be two

D

A

K

H

H

G

BC

E B

unequal Magnitudes,

whereof AB is the greater. I fay, if from AB be taken a greater Part than half, and from the Part

remaining

remaining there be again taken a Part greater than its half, and this be done continually, there will remain a Magnitude at last that shall be less than the Magnitude C.

For C being fome Number of Times multiplied, will become greater than the Magnitude AB. Let it be multiplied, and let DE be a Multiple of C greater than AB. Divide DE into Parts DF, FG, GE each equal to C, and take BH a Part greater than half of AB from AB, and again from AH the Part HK greater than half AH, and from AK a Part greater than half AK, and so on, until the Divifions that are in AB are equal in Number to the Divifions in DE. Therefore let the Divifrons AK, KH, HB, be equal in Number to the Divifions DF, FG, GE. Then because DE is greater than AB, and the Part EG is taken from ED, being lefs than half thereof, and the Part BH greater than half of AB is taken from it, the Part remaining GD, fhall be greater than the Part remaining HA. Again, because GD is greater than HA; and GF being half of GD, is taken from the fame; and HK being greater than half HA, is taken from this likewife; the Part remaining FD, fhall be greater than the Part remaining AK; but FD is equal to C. Therefore C is greater than AK; and fo the Magnitude AK is leffer than C. Therefore the Magnitude AK being the Part remaining of the Magnitude A B, is lefs than the leffer propos'd Magnitude C; which was to be demonftrated. If the Halves of the Magnitudes fhould 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.

ET ABCD, EFGH, be Circles, whofe Diameters are BD, FH. I fay, as the Square of BD is to the Square of FH, fo the Circle ABCD to the Circle EFGH.

For

* 41. 1.

For if it be not fo, the Sqaure of BD fhall be to the Square of FH, as the Circle ABCD is to fome Space either lefs or greater than the Circle EFGH. First let it be to a Space S, lefs than the Circle EFGH, and let the Square EFGH be described therein.

*

This Square EFGH will be greater than half the Circle EFGH; 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 defcribed about it. Therefore the Square EFGH is greater than half the Circle EFGH. Let the Circumferences EF, FG, GH, HE, be bifected in the Points K, L, M, N, and join EK, KF, FL, LG, GM, MH, HN, NE. Then each of the Triangles EKF, FLG, GMH, HNE, will be greater than one half of the Segment of the 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 Right Lines EF, FG, GH, HE be compleated, each of the Triangles EKF, FLG, GMH, HNE is half of each of the corresponding Parallelograms; but the Segment is lefs than the Parallelogram. Wherefore each of the Triangles EKF, FLG, GMH, HNE is greater than one half of the Segment of the Circle in which it ftands. Therefore if these 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 from it, and again from the Part remaining a Part greater than half be taken, and you do this continually; there will at laft remain a Magnitude that will be lefs 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 EFGH exceeds the Space S, and then the remaining Polygon EKFLGMHN fhall be greater than the Space S. Alfo defcribe the Polygon AXBOCPDR in the Circle ABCD, fimilar to the

2

A

Polygon