Book 11
23
Since then the angle ACB is an angle in the semicircle ACB, therefore the angle ACB is right. [III. 31]
23
Therefore the square on AB is equal to the squares on AC, CB. [I. 47]
23
Hence the square on AB is greater than the square on AC by the square on CB.
23
But AC is equal to LO.
23
Therefore the square on AB is greater than the square on LO by the square on CB.
23
If then we cut off OR equal to BC, the square on AB will be greater than the square on LO by the square on OR. Q. E. F.
PROPOSITION 24.
24
If a solid be contained by parallel planes, the opposite planes in it are equal and parallelogrammic.
24
For let the solid CDHG be contained by the parallel planes AC, GF, AH, DF, BF, AE; I say that the opposite planes in it are equal and parallelogrammic.
24
For, since the two parallel planes BG, CE are cut by the plane AC, their common sections are parallel. [XI. 16]
24
Therefore AB is parallel to DC.
24
Again, since the two parallel planes BF, AE are cut by the plane AC, their common sections are parallel. [XI. 16]
24
Therefore BC is parallel to AD.
24
But AB was also proved parallel to DC; therefore AC is a parallelogram.
24
Similarly we can prove that each of the planes DF, FG, GB, BF, AE is a parallelogram.
24
Let AH, DF be joined.
24
Then, since AB is parallel to DC, and BH to CF, the two straight lines AB, BH which meet one another are parallel to the two straight lines DC, CF which meet one another, not in the same plane; therefore they will contain equal angles; [XI. 10] therefore the angle ABH is equal to the angle DCF.
