What is the measure of $\overset{\LARGE{\frown}}{AB}$ in the figure below? 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
$\triangle OBC(isosceles):\measuredangle OCJ=\measuredangle OBJ=45^o\\
\measuredangle ADE = 90^o\\
\triangle DAJ(isosceles):\measuredangle DAJ = \measuredangle DJA\\
\measuredangle JAB = \measuredangle JCB = \frac{\measuredangle{JOB}}{2}$
 A: 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.
$$\implies \angle AOB=60^\circ$$
A: $Let BD = a ~and~ r = radius\\
AODJ~ is~ a ~square\\
BC = r\sqrt2\\
Power ~of~ Point~ Theorem :a(a+\sqrt2) = r^2 \rightarrow r^2-ar\sqrt{2}-a^2=0\\
\therefore r=\frac{a\sqrt{2}+-a\sqrt{6}}{2}\implies \frac{r}{a}=\frac{\sqrt{6}+\sqrt{2}}{2}\\
law ~of~ sines:\frac{r}{a}=\frac{sen(45+\frac{x}{2})}{sen(\frac{x}{2})}\\
\frac{\sqrt{6}+\sqrt{2}}{2}=\frac{\frac{\sqrt{2}}{2}(cos(\frac{x}{2})+sen(\frac{x}{2}))}{sen(\frac{x}{2})}\\
\sqrt{3}+1=\frac{(1+tg(\frac{x}{2}))}{tg(\frac{x}{2})}\\
k\sqrt{3}+k=1+k\implies k=\frac{\sqrt{3}}{3}\\
\therefore x = 60^o$
A: Using the annotated diagram in the question, let $B = (1, 0)$ and $C = (0, -1)$ on the coordinate plane such that $\angle BOC = 90º$. Also, suppose that $A = (\cos t, \sin t)$ which lies on the given circle.
Since $ADJO$ is a square, $D$ can be reached by adding the normal vector of the same length $(-\sin t, \cos t)$ to $A$, which is $(\cos t, \sin t) + (-\sin t, \cos t) = (\cos t - \sin t, \sin t + \cos t)$.
Now the intersection $D$ must lie on the line connecting B and C, which is just $y = x - 1$. Therefore $\sin t - \cos t = \cos t + \sin t - 1 \implies 2 \cos t - 1 = 0, t = 60º, 300º$ where $0 ≤ t < 360º$. However, due to the position of $A$, $0 ≤ t ≤ 90º$, so $t = 60º$ only.
Since $(\cos t, \sin t)$ always forms an angle $t$ with the $x$-axis, the measure of $AB$ is $\boxed{60º}$.
A: Comment: may be this figure helps users to find a solution.

