Kitap 11
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If a parallelepipedal solid be cut by a plane through the diagonals of the opposite planes, the solid will be bisected by the plane.
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For let the parallelepipedal solid AB be cut by the plane CDEF through the diagonals CF, DE of opposite planes; I say that the solid AB will be bisected by the plane CDEF.
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For, since the triangle CGF is equal to the triangle CFB, [I. 34] and ADE to DEH, while the parallelogram CA is also equal to the parallelogram EB, for they are opposite, and GE to CH, therefore the prism contained by the two triangles CGF, ADE and the three parallelograms GE, AC, CE is also equal to the prism contained by the two triangles CFB, DEH and the three parallelograms CH, BE, CE; for they are contained by planes equal both in multitude and in magnitude. [XI. Def. 10]
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Hence the whole solid AB is bisected by the plane CDEF. Q. E. D.
PROPOSITION 29.
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Parallelepipedal solids which are on the same base and of the same height, and in which the extremities of the sides which stand up are on the same straight lines, are equal to one another.
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Let CM, CN be parallelepipedal solids on the same base AB and of the same height, and let the extremities of their sides which stand up, namely AG, AF, LM, LN, CD, CE, BH, BK, be on the same straight lines FN, DK; I say that the solid CM is equal to the solid CN.
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For, since each of the figures CH, CK is a parallelogram, CB is equal to each of the straight lines DH, EK, [I. 34] hence DH is also equal to EK.
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Let EH be subtracted from each; therefore the remainder DE is equal to the remainder HK.
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Hence the triangle DCE is also equal to the triangle HBK, [I. 8, 4] and the parallelogram DG to the parallelogram HN. [I. 36]
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For the same reason the triangle AFG is also equal to the triangle MLN.
