geometry
For reference:
In the figure calculate $\overset{\LARGE{\frown}}{AB}$ if $ \overset{\LARGE{\frown}}{BC} = 90^o$
My progress:
Relationships I found:
$FO$ is angle bissector
$\triangle OBC(isosceles):\measuredangle OCJ=\measuredangle OBJ=45^o\\
\measuredangle ADE = 90^o\\
\triangle DAJ(isosceles):\measuredangle DAJ = \measuredangle DJA\\
\measuredangle JAB = \measuredangle JCB = \frac{\measuredangle{JOB}}{2}$
Let $B$ be a center of the third circle.
Thus, $B$, $G$ and $F$ are placed on the same line, $BG=BD$ and $FG=FE$.
Id est, $$x=\measuredangle DGB+\measuredangle EGF=\frac{1}{2}\left(\measuredangle ABF+\measuredangle AFB\right)=\frac{1}{2}\cdot140^{\circ}=70^{\circ}.$$
$\angle BAC + \angle FAE = 65^0$
$\angle PAQ = \frac{\angle POQ}{2} = 90^\circ - \frac{x}{2}$
$\angle BAF + \angle BDF = 180^\circ$
$\implies 65^\circ + 90^\circ - \frac{x}{2} + x = 180^\circ$
$\implies x = 50^\circ$
Best Answer
As it has been pointed out in comments, the information provided is not enough. Though according to your diagram, I assume $\small{ AD \perp DF}$ (according to the letters in my drawing)
Let $\small{\angle FOB=2x}$ so $\small{\angle FOE=x}$. And $\small{\angle BCF=x}$
You can clearly see $\small{OADF}$ is a square. Also $\small{\triangle OEF}$ and $\small{\triangle DFG}$ are equivalent. Therefore $\small{\angle FDG=\angle EOF=x}$.
Since $\small{\angle FCD=\angle FDG}$, $\small{\triangle CFD}$ is isosceles.$\small{\implies CF=DF}$
As $\small{OF=DF}$ (sides of the square) $\small{\triangle OCF}$ is equilateral ($\small{\because OF=OC=CF}$, first two as radii of the circle). Immediately you can prove $\small{\triangle OAB}$ is equilateral as well.
$$\implies \angle AOB=60^\circ$$