We have $\angle{FOA}=\angle{FBA}=\angle{CDA}=\angle{OCD}=\angle{OED}$ and $\angle{AFO}=\angle{ABO}=\angle{OAB}=\angle{DCE}=\angle{DOE}$. Hence $\triangle AOF \sim \triangle DEO$. Therefore $$\dfrac{AO}{DE}=\dfrac{AF}{DO}\implies{}R^2=AF\cdot{}DE.$$

Invert around $(O)$. Let $AB \cap CD$ $=$ $X$. Then $F, E$ invert to the midpoints of $AX$ and $DX$, from which applying Thales theorem and using the definition of inversion, we get the result.

We need to prove AF/AO = DO/DE. first we need to notice ∠FOA = 180 - ∠ABF = ∠CDA and ∠DOE = ∠DCE = DAB. AF/AO = sin ∠AOF / sin ∠ABO DO/DE = sin ∠DCO / sin ∠DCE sin ∠AOF / sin ∠DCO = sin ∠CDO / sin ∠DCO = CO/OD (1) sin ∠ABO / sin ∠DCE = sin ∠ABO / sin ∠OAB = AO/BO (2) from 1,2 we have AF/AO = DO/DE as we wanted so we're Done.