Then, since
DA is equal to
DB,
the angle DAE is also equal to the angle DBE. [I. 5] And, since one side
AEB of the triangle
DAE is produced,
the angle DEB is greater than the angle DAE. [I. 16]
But the angle
DAE is equal to the angle
DBE;
therefore the angle DEB is greater than the angle DBE. And the greater angle is subtended by the greater side; [
I. 19]
therefore DB is greater than DE. But DB is equal to DF;
therefore DF is greater than DE,
the less than the greater : which is impossible.
Therefore the straight line joined from
A to
B will not fall outside the circle.
Similarly we can prove that neither will it fall on the circumference itself;
therefore it will fall within.
Therefore etc. Q. E. D.
PROPOSITION 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.
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.
For let the centre of the circle
ABC
be taken, and let it be
E; let
EA,
EB be joined.
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]
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.
Therefore
CD, which is through the centre, and bisects
AB which is not through the centre, also cuts it at right angles.
Again, let
CD cut
AB at right angles;
I say that it also bisects it. that is, that
AF is equal to
FB.
