PROPOSITION 32.
If a straight line touch a circle,
and from the point of contact there be drawn across,
in the circle,
a straight line cutting the circle,
the angles which it makes with the tangent will be equal to the angles in the alternate segments of the circle.
For let a straight line
EF touch the circle
ABCD at the point
B, and from the point
B let there be drawn across, in the circle
ABCD, a straight line
BD cutting it; I say that the angles which
BD makes with the tangent
EF will be equal to the angles in the alternate segments of the circle, that is, that the angle
FBD is equal to the angle constructed in the segment
BAD, and the angle
EBD is equal to the angle constructed in the segment
DCB.
For let
BA be drawn from
B at right angles to
EF, let a point
C be taken at random on the circumference
BD, and let
AD,
DC,
CB be joined.
Then, since a straight line
EF touches the circle
ABCD at
B, and
BA has been drawn from the point of contact at right angles to the tangent, the centre of the circle
ABCD is on
BA. [
III. 19]
Therefore
BA is a diameter of the circle
ABCD;
therefore the angle ADB, being an angle in a semicircle, is right. [III. 31]
Therefore the remaining angles
BAD,
ABD are equal to one right angle. [
I. 32]
But the angle
ABF is also right; therefore the angle
ABF is equal to the angles
BAD,
ABD.
Let the angle
ABD be subtracted from each; therefore the angle
DBF which remains is equal to the angle
BAD in the alternate segment of the circle.
Next, since
ABCD is a quadrilateral in a circle, its opposite angles are equal to two right angles. [
III. 22]
But the angles
DBF,
DBE are also equal to two right angles; therefore the angles
DBF,
DBE are equal to the angles
BAD,
BCD,
of which the angle BAD was proved equal to the angle DBF; therefore the angle DBE which remains is equal to the angle DCB in the alternate segment DCB of the circle.
Therefore etc. Q. E. D.
