Book 3
3
But the right angle AFE is equal to the right angle BFE, therefore EAF, EBF are two triangles having two angles equal to two angles and one side equal to one side, namely EF, which is common to them, and subtends one of the equal angles; therefore they will also have the remaining sides equal to the remaining sides; [I. 26] therefore AF is equal to FB.
3
Therefore etc. Q. E. D. with the same construction, τῶν αὐτῶν κατασκευασθέντων.
PROPOSITION 4.
4
If in a circle two straight lines cut one another which are not through the centre, they do not bisect one another.
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.
