Sol.—Take any point A in the circumference. Draw the tangent AD, and make the angle DAC equal to the given angle X. AC will cut off the required segment.
Dem.—Take any point B in the alternate segment. Join BA, BC. Then the angle DAC is equal to ABC [xxxii.]; but DAC is equal to X (const.). Therefore the angle ABC is equal to X.
PROP. XXXV.—Theorem.
If two chords (AB, CD) of a circle intersect in a point (E) within the circle, the rectangles (AE.EB, CE.ED) contained by the segments are equal.
Dem.—1. If the point of intersection be the centre, each rectangle is equal to the square of the radius. Hence they are equal.
2. Let one of the chords AB pass through the centre O, and cut the other chord CD, which does not pass through the centre, at right angles. Join OC. Now because AB passes through the centre, and cuts the other chord CD, which does not pass through the centre at right angles, it bisects it [iii.]. Again, because AB is divided equally in O and unequally in E, the rectangle AE.EB, together with OE2, is equal to OB2—that is, to OC2 [II. v.]; but OC2 is equal to OE2 + EC2 [I. xlvii.] Therefore
Reject OE2, which is common, and we have AE.EB = EC2; but CE2 is equal to the rectangle CE.ED, since CE is equal to ED. Therefore the rectangle AE.EB is equal to the rectangle CE.ED.
