Kitap 3
5
For let the circles ABC, CDG cut one another at the points B, C; I say that they will not have the same centre.
5
For, if possible, let it be E; let EC be joined, and let EFG be drawn through at random.
5
Then, since the point E is the centre of the circle ABC, EC is equal to EF. [I. Def. 15]
5
Again, since the point E is the centre of the circle CDG, EC is equal to EG.
5
But EC was proved equal to EF also; therefore EF is also equal to EG, the less to the greater : which is impossible.
5
Therefore the point E is not the centre of the circles ABC, CDG.
5
Therefore etc. Q. E. D.
PROPOSITION 6.
6
If two circles touch one another, they will not have the same centre.
6
For let the two circles ABC, CDE touch one another at the point C; I say that they will not have the same centre.
6
For, if possible, let it be F; let FC be joined, and let FEB be drawn through at random.
6
Then, since the point F is the centre of the circle ABC, FC is equal to FB.
6
Again, since the point F is the centre of the circle CDE, FC is equal to FE.
6
But FC was proved equal to FB; therefore FE is also equal to FB, the less to the greater: which is impossible.
6
Therefore F is not the centre of the circles ABC, CDE.
6
Therefore etc. Q. E. D.
PROPOSITION 7.
