What's the measure of the segment $ON$ in the circle below?

Question

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 $AE (F=DC \cap AE)$

$\angle MAD = \angle AND\\ \triangle NOD \sim \triangle ADF \implies: \frac{OD}{ON} = \frac{FD}{AD}\implies ON = \frac{144\sqrt2}{FD}...$

Answer 1

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 $$|ON| =4.$$

Answer 2

We need only the simple picture:

$$ \frac{ON}{12} = \frac{ON}{OD} =\tan\delta=\tan(45^\circ-\alpha) =\frac{\tan 45^\circ -\tan \alpha}{1+\tan 45^\circ \tan \alpha} =\frac {1-(1/2)}{1+(1/2)}=\frac {1/2}{3/2}=\frac 13\ . $$ So $ON=4$.

$\square$

Answer 3

Please note that $\angle DEC = \angle DAC = 45^\circ$

So, $EN$ is angle bisector of $\angle AEC$. Also, $\triangle AEC \sim \triangle AOM$

So, $ \displaystyle \frac{AN}{NC} = \frac{AE}{EC} = \frac{AO}{OM} = 2$

Leads to $AN = 2 NC \implies NC = 8 $ and $ON = 4$