PROPOSITION 4. THEOREM.
If in a circle two straight lines cut one another, which do not both pass through the centre, they do not bisect one another.
Let ABCD be a circle, and AC, BD two straight lines in it, which cut one another at the point E, and do not both pass through the centre: AC, BD shall not bisect one another.
If one of the straight lines pass through the centre it is plain that it cannot be bisected by the other which does not pass through the centre.
But if neither of them pass through the centre, if possible, let AE be equal to EC, and BE equal to ED.
Take F the centre of the circle [III. 1.,
and join EF.
Then, because FE, a straight line drawn through the centre, bisects another straight line AC which does not pass through the centre; [Hypothesis.
FE cuts AC at right angles; ' [III. 3.
therefore the angle FEA is a right angle.
Again, because the straight line FE bisects the straight line BD, which does not pass through the centre, [Hyp.
FE cuts BD at right angles; [III. 3.
therefore the angle FEB is a right angle.
But the angle FEA was shewn to be a right angle;
therefore the angle FEA is equal to the angle FEB,[Ax.11.
the less to the greater; which is impossible.
Therefore AC, BD do not bisect each other.,
Wherefore, if in a circle &c. q.e.d.
PROPOSITION 5. THEOREM.
If two circles cut one another, they shall not have the same centre.
Let the two circles ABC, CDG cut one another at the
