Kitap 2
10
And EF is equal to CD; [I. 34] therefore the square on EG is double of the square on CD. But the square on EA was also proved double of the square on AC; therefore the squares on AE, EG are double of the squares on AC, CD.
10
And the square on AG is equal to the squares on AE, EG; [I. 47] therefore the square on AG is double of the squares on AC, CD. But the squares on AD, DG are equal to the square on AG; [I. 47] therefore the squares on AD, DG are double of the squares on AC, CD.
10
And DG is equal to DB; therefore the squares on AD, DB are double of the squares on AC, CD.
10
Therefore etc. Q. E. D.
Proposition 11.
11
To cut a given straight line so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment.
11
Let AB be the given straight line; thus it is required to cut AB so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment.
11
For let the square ABDC be described on AB; [I. 46] let AC be bisected at the point E, and let BE be joined; let CA be drawn through to F, and let EF be made equal to BE; let the square FH be described on AF, and let GH be drawn through to K.
11
I say that AB has been cut at H so as to make the rectangle contained by AB, BH equal to the square on AH.
11
For, since the straight line AC has been bisected at E, and FA is added to it, the rectangle contained by CF, FA together with the square on AE is equal to the square on EF. [II. 6]
11
But EF is equal to EB; therefore the rectangle CF, FA together with the square on AE is equal to the square on EB.
11
But the squares on BA, AE are equal to the square on EB, for the angle at A is right; [I. 47] therefore the rectangle CF, FA together with the square on AE is equal to the squares on BA, AE.
11
Let the square on AE be subtracted from each; therefore the rectangle CF, FA which remains is equal to the square on AB.
