+129852 Look at the following diagram, NSS......
Triangle DPC similar to triangle APB......therefore
DP/PC = AP/PB .... so.....
DP*PB = PA*PC
So.....the theorem is that when two secants are drawn from an external point outside a circle, the product of the length of one secant and its external segment equals the product of the length of the other secant and its external segment
Applying this to your problem we have that
AC*CB = (CD + x) * CD Note......AC = AB + BC
(AB + BC) * BC = (CD + x) * CD
(34 + 12) * 12 = (8 + x) * 8 simplify
(46)*12 = 64 + 8x
552 = 64 + 8x subtract 64 from both sides
488 = 8x divide both sides by 8
61 = x
[Looks as though you got the correct answer !!! ]
