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 equal the angles in the alternate segments of the circle, that is, that the angle FBD equals the angle constructed in the segment BAD, and the angle EBD equals the angle constructed in the segment DCB.
Draw BA from B at right angles to EF, take a point C at random on the circumference BD, and join AD, DC, and CB.
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 center of the circle ABCD is on BA.
Therefore BA is a diameter of the circle ABCD. Therefore the angle ADB, being an angle in a semicircle, is right.
Therefore the sum of the remaining angles BAD and ABD equals one right angle.
But the angle ABF is also right, therefore the angle ABF equals the sum of the angles BAD and ABD.
Subtract the angle ABD from each. Therefore the remaining angle DBF equals the angle BAD in the alternate segment of the circle.
Next, since ABCD is a quadrilateral in a circle, the sum of its opposite angles equals two right angles.
But the sum of the angles DBF and DBE also equals two right angles, therefore the sum of the angles DBF and DBE equals the sum of the angles BAD and BCD, of which the angle BAD was proved equal to the angle DBF, therefore the remaining angle DBE equals the angle DCB in the alternate segment DCB of the circle.
Therefore if a straight line touches a circle, and from the point of contact there is drawn across, in the circle, a straight line cutting the circle, then the angles which it makes with the tangent equal the angles in the alternate segments of the circle.