1.The Chord Intersection Theorem:
If 2 chords of a circle are AB and CD and they intersect at E, then
AE * EB = CE * ED.
Problem.
Two Chords AB and CD intersect  at E.  If AE =  2cm , EB = 4 and CE = 2.5 cm, find the length of ED.
By the above theorem : 2 * 4 = 2.5 * ED
ED = (2 * 4) / 2.5
= 3.2 cm.
2.1. A given chord on a circle is perpendicular to a radius through its center, and it is at a distance less that the radius of the circle.
2. A circle of center O has a radius of 13 units. If a chord AB of 10 units is drawn at a distance, d, to the center of the circle, determine the value of d.
3. From question 2, the radius = 13 units, length of chord = 10 units and distance of chord to center of the circle is d.
A radius that meet the chord at center C, and divides it into two equal parts.
So that;
AC = CB = 5 units
Applying Pythagoras theorem to ΔOCB,
OC = d, CB = 5 units and OB = 13 units
= Â + Â
169 = 25 + Â
169 - 25 = Â
144 = Â
⇒ d = Â
   = 12 units
Therefore, the chord is at a distance of 12 units to the center of the circle
