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Proposition 24

If a solid is contained by parallel planes, then the opposite planes in it are equal and parallelogrammic.
Let the solid CDHG be contained by the parallel planes AC, GF, AH, DF, BF, and AE. I say that the opposite planes in it are equal and parallelogrammic.
| Since the two parallel planes BG and CE are cut by the plane AC, therefore their common sections are parallel. Therefore AB is parallel to DC. Again, since the two parallel planes BF and AE are cut by the plane AC, therefore their intersections are parallel. Therefore BC is parallel to AD. | XI.16
But AB was proved parallel to DC, therefore AC is a parallelogram. Similarly we can prove that each of the planes DF, FG, GB, BF, and AE is a parallelogram. Join AH and DF.
Then, since AB is parallel to DC, and BH is parallel to CF, therefore the two straight lines AB and BH, which meet one another, are parallel to the two straight lines DC and CF, which meet one another, not in the same plane. Therefore they contain equal angles. Therefore the angle ABH equals the angle DCF. | XI.10
And, since the two sides AB and BH equal the two sides DC and CF, and the angle ABH equals the angle DCF, therefore the base AH equals the base DF, and the triangle ABH equals the triangle DCF. | I.34 I.4
And the parallelogram BG is double the triangle ABH, and the parallelogram CE is double the triangle DCF, therefore the parallelogram BG equals the parallelogram CE. | I.34
Similarly we can prove that AC equals GF, and AE equals BF.
Therefore, if a solid is contained by parallel planes, then the opposite planes in it are equal and parallelogrammic.
Q. E. D.

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This proposition is used in the next as well as others in this book and the next.

Book XI Introduction - Proposition XI.23 - Proposition XI.25.

© 1996
D.E.Joyce
Clark University