Kitap 3
14
And, since AE is equal to EC, the square on AE is also equal to the square on EC. But the squares on AF, EF are equal to the square on AE, for the angle at F is right; and the squares on EG, GC are equal to the square on EC, for the angle at G is right; [I. 47] therefore the squares on AF, FE are equal to the squares on CG, GE, of which the square on AF is equal to the square on CG, for AF is equal to CG; therefore the square on FE which remains is equal to the square on EG, therefore EF is equal to EG.
14
But in a circle straight lines are said to be equally distant from the centre when the perpendiculars drawn to them from the centre are equal; [III. Def. 4] therefore AB, CD are equally distant from the centre.
14
Next, let the straight lines AB, CD be equally distant from the centre; that is, let EF be equal to EG.
14
I say that AB is also equal to CD.
14
For, with the same construction, we can prove, similarly, that AB is double of AF, and CD of CG.
14
And, since AE is equal to CE, the square on AE is equal to the square on CE. But the squares on EF, FA are equal to the square on AE, and the squares on EG, GC equal to the square on CE. [I. 47]
14
Therefore the squares on EF, FA are equal to the squares on EG, GC, of which the square on EF is equal to the square on EG, for EF is equal to EG; therefore the square on AF which remains is equal to the square on CG; therefore AF is equal to CG. And AB is double of AF, and CD double of CG; therefore AB is equal to CD.
14
Therefore etc. Q. E. D.
PROPOSITION 15.
15
Of straight lines in a circle the diameter is greatest, and of the rest the nearer to the centre is always greater than the more remote.
15
Let ABCD be a circle, let AD be its diameter and E the centre; and let BC be nearer to the diameter AD, and FG more remote; I say that AD is greatest and BC greater than FG.
15
For from the centre E let EH, EK be drawn perpendicular to BC, FG.
15
Then, since BC is nearer to the centre and FG more remote, EK is greater than EH. [III. Def. 5]
