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PROPOSITION XXVIII.

34. I.

+ THEOREM. If a solid Parallelepipedon be cut by a Plone paf

fing thro' tbe Diagonals of two opposite Planes,

that Solid will be biseEted by the Plane. LET the folid Parallelepipedon A B be cut by the

Plane CDEF, passing thro' the Diagonals CF, DE, of two opposite Planes. I say, the Solid A B is bisected by the Plane CDEF.

For, because the Triangle C G F is * equal to the

Triangle CBF, and the Triangle ADE to the Triangle +240 tbis. D EH, and the Parallelogram C A to + the Parallelo

gram B E, for it is opposite to it; and the Parallelogram

GE to the Parallelogram CH; the Prism contained by the two Triangles CGF, A DE, and the three Parallelograms GĚ, AC, CE, is equal to the Prism contained under the two Triangles CFB, DEH,

and the three Parallelograms CH, BE, CE; for I they Def. 10. of rbis,

are contained under Planes equal in Number and Magnitude. Therefore, the whole Solid A B is bifected by the Plane CDEF; wbich was to be demonstrated.

PROPOSITION XXIX. .

THE OʻREM.
Solid Parallelepipedons, being constituted upon the

fame Base, and having the same Altitude, and
wbose insistent Lines are in the same Right

Lines, are equal to one another.
L
ET the folid Parallelepipedons CM, B F, be

constituted upon the same Bale AB, with the fame Altitude, whose infiftent Lines AF, AG, LM, LN, CD, CE, BH, BK, are in the same Right Lines FN, DK. I say, the Solid CM is equal to the Solid BF.

For, because CH, CK, are both Parallelograms, C B Ball be * equal to D H, or E K; wherefore DH

is

34. 1.

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is equal to E K, Let E H, which is common, be taken
away, then the Remainder D E will be equal to the
Remainder H K, and so the Triangle D E Cis t equal + 8. 1.
to the Triangle HK B, and the Parallelogram D G
equal to the Parallelogram HN; for the fame Reason
the Triangle AF G is equal to the Triangle L MN.
Now the Parallelogram CF I is equal to the Paralle- 1 24 of tibien
logram BM, and the Parallelogram C G to the Paral-
lelogram BN, for they are opposite. Therefore the
Prilin contained under the two Triangles AFG,
DE C, and the three Parallelograms CF, DG,CG,
is * equal to the Prism contained under the two Tri-* Def. o.
angles L MN, HB K, and the three Parallelogramsef ibis,
BM, HN, B N. Let the common Solid, whose Base
is the Parallelogram AB, oppofite to the Parallelo-
gram GEHM, be added, then the whole solid Paral-
lelepipedon CM is equal to the whole solid Parallele-
pipedon BF. Therefore, folid Parallelepipedons, being
conflitued upon the same Base, and having the same Alti-
tude, and whose insistent Lines are in the fame Right Lines,
are equal to one another; which was to be demonstrated.

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Solid Parallelepipedons, being constituted upon the
same Base, and having the fame Altitude, whose
insistent Lines are not placed in the same Right

Lines, are equal to one another.
LET there be solid Parallelepipedons CM, CN:

having equal Altitudes, and standing on the same
Base A B, and whose infiftent Lines AF, AG, LM,
LN,CD, CE, BH, BK, are not in the same Right
Lines. I say, the Solid C M is equal to the Sotid
CN.

For let NK, DH, be produced, and G E, FM; be drawn, meeting each other in the Points R, X: Let also FM, GE, be produced to the Points O, P, and join AX,LO, CP, BR. The Solid CM, whofe Bart is the Parallelogram A CB L, being opposite to the Parallelogram FDHM, is * equal to the Solid CO, 29 of thien 6

whole

whose Base is the Parallelogram ACBL, being opposite to XPRO, for they stand upon the fame Bare ACBL; and the insistent Lines AF, AX, LM, LO,CD, CP, BH, BR, are in the same right Lines FO,DR:But the Solid CO, whose Base is the Pa

rallelogram ACB L, being opposite to X PRO, is * 390f idis. * equal to the Solid CN, whose Base is the Paralle

logram ACBL, being opposite to GEKN; for they stand upon the same Base A CB L, and their insistent Lines AG, AX, CE, CP, LN, LO, BKBR, are in the fame Right Lines GP, NR: Wherefore the Solid C M shall be equal to the Solid CN. Therefore, solid Parallelepipedons, being constituted upon the fame Base, and having the same Altitude, whose infiftent Lines are not placed in the same Right Lines, are equal to one another ; which was to be demonstrated.

PROPOSITION XXXI.

THEOREM.

Solid Parallelepipedons, being constituted upon

equal Bases, and baving the fame Altityde, are

equal to one another. LE

ETAE, CF, be folid Parallelepipedons, consti

tuted upon the equal Bases A B, C D, and having the fame Altitude. I say, the Solid A E is equal to the Solid CF.

First, Let HK, BE, AG, LM, OP, DF, C., RS, be at Right Angles to the Bafes A B, CD; let the Angle A L B not be equal to the Angle CRD, and produce CR to T, so that R T be equal te AL, then

make the Angle TRY, at the Point R, in the Right * 23. 1. * Line RT, equal * to the. Angle ALB; make RRY

equal to L B; draw X Y, thro' the Point Y, * parallel $31.1. to RT I, and compleat the Parallelogram R X, and

the Solid Y Y. Therefore, because the two Sides TR, RY, are equal to the two Sides A L, L B, and they contain equal Angles; the Parallelogram R X shall be equal and similar to the Parallelogram HL. . And again, because A L is equal to RT, and L M to RS, and shey contain equal Angles, the Parallelogram Ry

thall

.

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