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
$\triangle OBC(isosceles):\measuredangle OCJ=\measuredangle OBJ=45^o\\
\measuredangle ADE = 90^o\\
\triangle DAJ(isosceles):\measuredangle DAJ = \measuredangle DJA\\
\measuredangle JAB = \measuredangle JCB = \frac{\measuredangle{JOB}}{2}$

$Let BD = a ~and~ r = radius\\
AODJ~ is~ a ~square\\
BC = r\sqrt2\\
Power ~of~ Point~ Theorem :a(a+\sqrt2) = r^2 \rightarrow r^2-ar\sqrt{2}-a^2=0\\
\therefore r=\frac{a\sqrt{2}+-a\sqrt{6}}{2}\implies \frac{r}{a}=\frac{\sqrt{6}+\sqrt{2}}{2}\\
law ~of~ sines:\frac{r}{a}=\frac{sen(45+\frac{x}{2})}{sen(\frac{x}{2})}\\
\frac{\sqrt{6}+\sqrt{2}}{2}=\frac{\frac{\sqrt{2}}{2}(cos(\frac{x}{2})+sen(\frac{x}{2}))}{sen(\frac{x}{2})}\\
\sqrt{3}+1=\frac{(1+tg(\frac{x}{2}))}{tg(\frac{x}{2})}\\
k\sqrt{3}+k=1+k\implies k=\frac{\sqrt{3}}{3}\\
\therefore x = 60^o$
