In $\Delta ABC$ , $\angle A = 20^\circ, \angle C = 90^\circ$. $O$ is a point on $AB$ and $D$ is the midpoint of $OB$ . Find $\angle BCD$ . 
In $\Delta ABC$ , $\angle A = 20^\circ, \angle C = 90^\circ$. $O$ is a  point on $AB$ and $D$ is the midpoint of $OB$ . The circle centered at $O$ with radius $OD$ intersects $AC$ at $T$. Find $\angle BCD$ .

What I Tried: Here is a picture :-

You can see what I tried. The first part is to do angle-chasing and find the respective angles in the picture. I also find that $OT \parallel BC$, since $OBCT$ is a trapezium . Now I think that to find $\angle BCD$ , somehow I have to use the fact that $OD = DB$ (which I have not used) and maybe the parallel lines, but I seem to be stuck here.
Can anyone help me figure this out? Thank You.
 A: Construct $TD$ and a line parallel to $OT$ passing through $D$.

By Intercept theorem, $TO// GD//CB$ and $OD = DB$ implies $TG = GC$.
Since $GD$ is common side and $TC \perp GD$, $\triangle TGD \cong \triangle CGD$ via SAS. This gives $\angle TDG = \angle CDG$.
Now $\angle TDO = \frac12 \angle TOA$ by circle properties and $\angle GDA = \angle TOA$ by parallel lines, giving $$\angle TDO = \angle TDG = \angle CDG = \frac12 \angle TOA$$ This gives $$\angle CDO = \frac32 \angle TOA$$
Finally by considering the exterior angle of $\triangle BCD$, $$\angle BCD + \angle CBD = \angle CDO = \frac32 \angle TOA$$
but $\angle CBA = \angle TOA$, giving $\angle BCD = \frac12 \angle TOA$.
In this case, it is equal to $\frac12(70^\circ) = 35^\circ$.
A: Given you already have a geometric solution, here is a trigonometric one.
Say $\angle BCD = x, $ radius of the circle = $r$. Portion of $BA$ on the left outside circle is say, $a$.
Using Sine law in $\triangle BCD$,
$\displaystyle \frac{r}{\sin x} = \frac{CD}{\sin 70^0}$ ...(i)
(as $BD = r$)
Applying Sine law in $\triangle AOT$,
$\displaystyle \frac{a+r}{\sin 90^0} = \frac{r}{\sin 20^0} \implies a = \frac{r(1-sin 20^0)}{\sin 20^0}$ ...(ii)
(as $AO = a + r, OT = r$)
Applying Sine Law in $\triangle ACD$,
$\displaystyle \frac{a+2r}{\sin (90^0-x)} = \frac{CD}{\sin 20^0}$
(as $AD = a + 2r, \angle ACD = (90^0 - x)$)
So, $\displaystyle \frac{a+2r}{\cos x} = \frac{CD}{\sin 20^0} \implies CD = r\frac{1 + \sin20^0}{\cos x}$ (substituting $a$ from (ii))
Now substituting $CD$ in (i),
$\tan x = \displaystyle \frac{\sin 70^0}{1 + \sin 20^0} = \frac{\sin 70^0}{1 + \cos 70^0}$
i.ie $\tan x = \displaystyle \frac{2 \sin 35^0 \cos 35^0}{2 \cos^2 35^0} = \tan 35^0$
