Book 3
4
Let ABCD be a circle, and in it let the two straight lines AC, BD, which are not through the centre, cut one another at E; I say that they do not bisect one another.
4
For, if possible, let them bisect one another, so that AE is equal to EC, and BE to ED; let the centre of the circle ABCD be taken [III. 1], and let it be F; let FE be joined.
4
Then, since a straight line FE through the centre bisects a straight line AC not through the centre, it also cuts it at right angles; [III. 3] therefore the angle FEA is right.
4
Again, since a straight line FE bisects a straight line BD, it also cuts it at right angles; [III. 3] therefore the angle FEB is right.
4
But the angle FEA was also proved right; therefore the angle FEA is equal to the angle FEB, the less to the greater: which is impossible.
4
Therefore AC, BD do not bisect one another.
4
Therefore etc. Q. E. D.
PROPOSITION 5.
5
If two circles cut one another, they will not have the same centre.
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.
