Theorem: If two chords intersect within a circle, then the product of the lengths of the segments (parts) of one chord is equal to the product of the lengths of the segments of the other chord.
In your case,
[tex]AE\cdot EB=CE\cdot ED.[/tex]
If AE = 6, EB = 3, and DE = 2, then
[tex]6\cdot 3=CE\cdot 2,\\ \\CE=\dfrac{6\cdot 3}{2}=9.[/tex]
Answer: correct choice is D.
