Hello dgen,

This is what we do. There's a theorem out there that states: If two secants meet at a point outside the circle, then the product of the outside segment and secant of the first is equal to the product of the outside segment and secant of the second. So, based on your diagram:

ED(EC) = (EA)(AB)

ED = 8, DC = 4. Therefore EC = 12

EA = 6

8(12) = 6(AB + 6)

96 = 6(AB) + 36

60 = 6(AB)

10 = AB