Book 3
2
Therefore etc. Q. E. D.
PROPOSITION 3.
3
If in a circle a straight line through the centre bisect a straight line not through the centre, it also cuts it at right angles; and if it cut it at right angles, it also bisects it.
3
Let ABC be a circle, and in it let a straight line CD through the centre bisect a straight line AB not through the centre at the point F; I say that it also cuts it at right angles.
3
For let the centre of the circle ABC be taken, and let it be E; let EA, EB be joined.
3
Then, since AF is equal to FB, and FE is common, two sides are equal to two sides; and the base EA is equal to the base EB; therefore the angle AFE is equal to the angle BFE. [I. 8]
3
But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [I. Def. 10] therefore each of the angles AFE, BFE is right.
3
Therefore CD, which is through the centre, and bisects AB which is not through the centre, also cuts it at right angles.
3
Again, let CD cut AB at right angles; I say that it also bisects it. that is, that AF is equal to FB.
3
For, with the same construction, since EA is equal to EB, the angle EAF is also equal to the angle EBF. [I. 5]
