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E

is in the fame plane CD; [by

с.
3. def. 11.) the angle FG E
is a right angle. But e Falfo is
at right angles to the plane
AB
Wherefore E F G is a Al

D
right angle: and so two an-
gles of the triangle & F G are
right angles; which{by 17.1.)
is impoñble: Wherefore a perpendicular drawn from the
point e to the plane A B does not fall out of the right line
AD: Therefore it falls in AD.

Wherefore if one plante be at right angles to another, a right line drawn from any point in one plane perpendicular to the other, will fall in the common soction of the planes. Which was to be demonstrated.

PROP. XXXIX. THEOR.
If the fides of the opposite planes of a solid parallele-

pipedon be biseEted, and planes be drawn tbro the
sections ; the common seation of the planes; and the
diameter of the folid parallelepidon, will mutually
bi seat each other.

For let the fides of the opposite planes C F, A # of the
parallelepipedon Dy be bisected in the points K, L; M,N
X, P, 0, 0, and through the sections draw the planes KN,
xQ, and let y s be the common fection of the planes, and
DG the diameter of the solid parallelepipedon: I say Y-s,
DO da mutually cut one another into two equal parts, that
is, Yt is equal to T-Sg
and D T equal to TG,

D
For join D Y, YE,
BS, SG: Then be-

L
cause D x is parallel to
OE; the alternate
angles DxY, YOE
[by: 29. r.] are equal
to one another; and
because D x is equal

M
to o E, and X Y to yo,

B В and they contain equal P

S angles; the base D Y

[by

K

T

by 4. 1.] will be equal to the base ve and the triangle DX Y is equal to the triangle y o E, and the rest of the angles equal to the rest of the angles: Therefore the an gle XYD is equal to the angle o Y E; and so [by 14. 1.] DY E is one right line. By the same reason Bs G is one right line, and b s is equal to 's G. And because by [34. 1.} c A is equal and parallel to D B, and c A is also equal and parallel to E G; DB will be equal to E G, and (by 30. 1.) parallel to it. But the right lines DE, BG join them: Therefore [by 33. 1] De is parallel to BG, and in cach of them some points D, Y, G, s, are taken, and DG, Y S are joined: Therefore [by 7. 11.] DG, Y S are both in one plane. Wherefore because Deis parallel to BG, the angle EDT [by 29. 1.] will be equal to the angle BGT; for they are alternate angles. But the angle DTY (by 15. 1.] is equal to the angle GTs; Therefore there are two triangles DT Y, GTS, having two angles of the one equal to two angles of the other, each to each, and one side of the one equal to one side of the other, opposite to one of the equal angles, viz. Dr.equal to GS; for they are the halves of DE, BG: Therefore (by 26. 1.] the rest of the sides will he equal to the rest of the fides; and foot is equal to TG, and IT equal to TS.

If therefore in a solid parallelepipedon, &c. Whịch was to be demonstrated,

PROP. XL. THEOR.
If there be two prisms of the same altitude, and the

base of one of them be a parallelogram, and
that of the other a triangle, and that parallelo-
gram be double to the triangle; the prisms will
be equal.

Let A B C D E F, G HKLMN be prisms of the same altitude. Let the base of one of them be the parallelogram AF, and that of the other the triangle G HK, and let the parallelogram A F be double to the triangle GHK: I say the prism A B C D E F is equal to the prism GH Ķ L MN.

For

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And M

For complete the solids AX, G 0.

B

M; because the parallelogram A F is double to

N

X the triangle GHK,

C and [by 34. 1.] the H

A parallelogram H K is double to the triangle

G K GHK; the parallelogram A F will be equal to the parallelogram hk. But (by 31. 11.) parallelepipedons standing upon equal bases, and having the same altitude, are equal to one another : Therefore the solid ax is equal to the solid Go.

But the prism A B C D E F is one half the solid ax, and the prism GHK L M N one half the folid Go: Therefore the prism ABC D E F is equal to the prism G H K L M.

If therefore two prisms, &c. Which was to be demonstrated,

EUCLID's

EU CLI D's ELEMENTS.

BOOK XII.

L E M M A. If there be two unequal magnitudes proposed, and

from the greater be taken away more than one half of it, and again from the remainder be taken away more than its half, and this be always done ; tbere will at last remain some magnitude that will be less than the least of the two given magnitudes.

L

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ET there be two unequal magnitudes A B, C, whereof

A B is the greater: I say it from A B be taken away more than one half, and from the remainder be again taken away more than its half, and this be always done, there will at last remain some magnitude lesser than c.

For if c be multiplied, it will fome time or other become greater than the magnitude A B. Let this be done, and let D E be a multiple of c greater than A B. Divide DE into the parts DF, FG, GE, each equal to c, and from A B take away BH more than its half, and again from AH take away hk more than one half of it, and do thus continually until there are the same number of divisions in A B as in D E. Let then these divisions A K, KH, HB, be equal in number to the divisions G F, FG, GE.

And because D E is greater than a B, and there is taken away from de less than one half of it, viz. EG, and from A B more than its half, viz. B H; the remainder GD will be greater than the remainder A H. Again, becaufe

2

GD

K

GD is greater than HA, and from GD

A is taken its half GF, and from HA more than its half HK, the remainder DF will be

greater than the remainder AK. But Df is equal to c: There

H! fore c is greater than AK: Wherefore AK is less than.c: Therefore the remainder AK of the magnitude AB, is less than the lesser of the two given magnitudes c. Which was to be demonftrated.

After the same manner we demonstrate the thing, when the halves of the given magnitudes are taken away.

Otherwise. Let A b, c be two given magnitudes, and let c'be the leaft of them. Then because c is the least, if it be multiplied, it will at length be greater than the magnitude A B. Let this be done, and let FM be the magnitude, which divide into the parts MH, HG, G F, each equal to c; and take away B E from A B, greater than one half of it; and from E A, take E D 'greater than one half of it; and doing thus continually until the number of divisions in AB are equal to the number of divifions in FM. Let these be BE, ED, DA; and KL, LN, NX, each equal to DA; and do this continually till the divisions of Kx be equal in number to the divisions in FM.

Then because be is greater than one half AB, BE will be greater

D+ than E A; therefore B E is much

X greater than DA, but xn is equal to DA: wherefore B E is greater

N than xn. Again, because E D is great

H Η er than one half È A; ED will be

L greater than DA. But nl is equal to DA: wherefore Ed is greater than BCM K К NL: therefore the whole is greater than xl: But Lk is equal to DA. Therefore the whole A B will be greater than the whole x K. But MF is greater than BA: Wherefore M

is much greater than xk And because x N, NL, L-k are equal to one

another,

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DB

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