SOLUTION: in circle o, chords AB and CD intersect at E. if AE=10, EB=6 & CE is 2 more than twice ED, find CE and ED

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Let ED = x.
Then CE = 2x + 2
According to the chord segment theorem in geometry, (CE)(DE)=(AE)(BE)
By substitution, we have

Simplifying, we get

dividing both sides by 2
Re-write it in standard form, we have:

Solving for x, we have

So x = 5 or x= -6 (reject this negative number)
Thus DE = 5. Therefore CE =2x + 2 = 12.