Kitap 11
23
Next, since by hypothesis the square on OR is equal to that area by which the square on AB is greater than the square on LO, therefore the square on AB is equal to the squares on LO, OR.
23
But the square on LR is equal to the squares on LO, OR, for the angle LOR is right; [I. 47] therefore the square on AB is equal to the square on RL; therefore AB is equal to RL.
23
But each of the straight lines BC, DE, EF, GH, HK is equal to AB, while each of the straight lines RM, RN is equal to RL; therefore each of the straight lines AB, BC, DE, EF, GH, HK is equal to each of the straight lines RL, RM, RN.
23
And, since the two sides LR, RM are equal to the two sides AB, BC, and the base LM is by hypothesis equal to the base AC, therefore the angle LRM is equal to the angle ABC. [I. 8]
23
For the same reason the angle MRN is also equal to the angle DEF, and the angle LRN to the angle GHK.
23
Therefore, out of the three plane angles LRM, MRN, LRN, which are equal to the three given angles ABC, DEF, GHK, the solid angle at R has been constructed, which is contained by the angles LRM, MRN, LRN. Q. E. F.
23
Lemma. But how it is possible to take the square on OR equal to that area by which the square on AB is greater than the square on LO, we can show as follows.
23
Let the straight lines AB, LO be set out, and let AB be the greater; let the semicircle ABC be described on AB, and into the semicircle ABC let AC be fitted equal to the straight line LO, not being greater than the diameter AB; [IV. 1] let CB be joined
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.
