« ZurückWeiter »
+ 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. .
fame Base, and having the same Altitude, and
Lines, are equal to one another.
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 equal to E K, Let E H, which is common, be taken
Solid Parallelepipedons, being constituted upon the
Lines, are equal to one another.
having equal Altitudes, and standing on the same
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
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.
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