CB = sqrt(CD^2 +(DE+EB)^2)
I drew up what I was asking.. I got there without angle B, and derived that first.. but threw it in there to assume anyway. You can also use the length of the chord if you wish.. just looking for the simplest method. I got there in several steps, and it's a bit of a mess.
That wasn't really the point of the thread. I asked for people willing to do so to provide the simplest solve they can, not I can.
As my solve was;
line CA = 2CBSin(CBA/2) / Sin90 (so that you can see I split triangle CBA down the middle).
angle ACB = Sin^-1(ABSinCBA / CA)
angle CAD = CAB = ACB
angle ACD = 180 - 90 - CAD
AD = ACSinACD / Sin90
DE = AB - AD - EB