What's the measure of the segment $ON$ in the circle below? For reference:On the circle with center $O$ and perpendicular diameters
perpendiculars $AC$ and $BD$ a string $AE (E~ in~ \overset{\LARGE{\frown}}{BC})$ is drawn
such that AE intersects $BD$ at $M$; if $BM = MO$. Calculate $ON$, if $OA = 12$. ($N =ED\cap AC$).
My progress:

$\triangle ABO(right): AB = 12\sqrt2 = AD\\\triangle ABM \sim \triangle DME (A.A)\\
\frac{12\sqrt2}{DE} = \frac{BM}{ME} = \frac{AM}{DM}\\
\angle ABD = \angle BEA = \angle AED =45^o\\
\angle DBE = \angle OND \implies\\
\frac{ON}{BE}=\frac{DN}{24}=\frac{6}{DE}$
Extend $DC$ and $AE (F=DC \cap AE)$
$\angle MAD = \angle AND\\
\triangle NOD \sim \triangle ADF \implies: \frac{OD}{ON} = \frac{FD}{AD}\implies ON = \frac{144\sqrt2}{FD}...$

Denote $\angle OAM =\alpha, \angle EAB =\beta.$
We have $\tan\alpha =\frac{|MO|}{|OA|} =\frac{1}{2}$ thus $\cos\alpha =2\sin\alpha$ and hence $$\sin\alpha =\frac{\sqrt{5}}{5} , \cos\alpha =\frac{2\sqrt{5}}{5}.$$
Now $$\sin\beta =\sin (\frac{\pi }{4} -\alpha ) =\frac{\sqrt{2}}{2}\cdot \frac{2\sqrt{5}}{5} -\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{5}}{5}=\frac{\sqrt{10}}{10},\cos\beta = \frac{3\sqrt{10}}{10}$$
Now it is easy to observe that $\angle BDE =\beta$ and hence
$$\frac{|ON|}{|OD|} =\tan\beta =\frac{1}{3}$$
therefore $$|ON| =4.$$
