Consider a trapezoid $ABCD,$ with major basis $AB,$ circumscribed to a circle of radius $R.$ Let $F$ be the intersection of the lines $AD$ and $BC.$ Choose the point $E$ on the line $CD,$ on the side of $C,$ such that the two following conditions are satisfied:
1) $$AF=2BE$$ 2) If $r$ is the radius of the circle inscribed in the triangle $BEC,$ then $R=2r.$
Then $CDF$ and $CBE$ are similar.
The statement is obvious if $BE$ and $DF$ are parallel, but I don’t know how to continue if they are not, in particular where to use the additional hypothesis that the two sides are not parallel (the situation in the figure below).
Clearly $DCF \sim ABF$, so it is enough to prove $ABF\sim CBE$.
Let $I$ be an incenter of $ABF$ and $J$ an incenter of $CBE$ and let perpendicular to $AF$ ($BE$) thorugh $I$ ($J$) cut $AB$ ($BC$) at $L$ ($M$).
Since $\angle ABF = \angle BCE$ we have $$\angle AIF = 90^{\circ}+{1\over 2}\angle ABF = 90^{\circ}+{1\over 2}\angle BCE=\angle BJE$$
Then $AFI \sim BEJ$ with coefficent of similarity $k=2$. (since $R=2r$ and $AF = 2BE$ and $\angle AIF = \angle BJE$). So $$\angle IAF = \angle JBE \;\;\;{\rm and}\;\;\;\angle IFA = \angle JEB$$
so $$\angle FAB =2\angle IAF = 2\angle JBE= \angle EBC $$vand
$$\angle BFA =2\angle IFA = 2\angle JEB = \angle CEB$$ and thus a conclusion.