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.
Best Answer
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 $\frac12(70^\circ) = 35^\circ$.