The following figure illustrates the given situation, and we need to show that  $$\frac{AB}{CD}=\frac{OA.OB}{OC.OD}$$

Observe that $$\frac{AB}{CD}=\frac{ar(OAB)}{ar(OCD)}=\frac{\frac{1}{2}OA\cdot OB\sin(\angle AOB)}{\frac{1}{2}OC\cdot OD\sin(\angle COD)}$$

Hence if $$\angle AOB=\angle COD$$ then we are done.
Now observe that $$\angle EOC+\angle COD=\angle EOD=\angle OAD=\angle OBA+\angle AOB=\angle EOC+\angle AOB$$
Hence we have $$\angle AOB=\angle COD$$

Here you should consider directed angles to avoid configuration issues