For part $a)$ By Pythagoras' Theorem in $\Delta OMA$, It's easy to find that, $\boxed{BM=R}$ Let $\angle AOB =x$, By Sine Rule in $\Delta OAB$ and $\Delta ABM$, $$\frac{R}{\cos \tfrac{x}{2}} =\frac{AB}{\sin x } $$$\text{and,}$ $$\frac{R\sqrt{3}}{\cos \tfrac{x}{2}} = \frac{R}{\sin \tfrac{x}{2}} = \frac{AB}{\cos x} $$ From which we get that, $$\frac{\cos \tfrac{x}{2}}{\sin x } = \frac{\sin \tfrac{x}{2}}{\cos x } \implies \tan x \cdot \tan \frac{x}{2} =1 $$Let $\tan \frac{x}{2} =t \implies t=\frac{1-t^2}{2t} \implies t=\frac{1}{\sqrt{3}}$, Hence, $\boxed{\angle AOB =x = 60^{\circ}}$ Therefore, $\angle ADM = \angle AMD = 30^{\circ}$ Hence, $\Delta AMD$ is Isosceles Triangle

For part $b)$ Let the $E$ be a point on segment $AD$ such that, $AE=ED$ and, since, $\Delta AOD$ is isosceles triangle, $OE$ is the perpendicular bisector of $AD$ Let $OE \cap C = F$ We have, $$\angle BOF = 60^{\circ} ; \angle AFE = 30^{\circ} , \angle BFA = 30^{\circ} \implies \angle BFO =60^{\circ} \implies \Delta BFO \text{is equilateral and,} BFOA \text{is Rhombus}$$ Let $AF \cap BO = G $ which implies that, $G$ is mid-point of side $BO$ Since, $BM=OD \implies GM=GD$ and Since, Diagonals intersect perpendicularly in a Rhombus Hence, $AG$ is the perpendicular bisector of length $DM$ and Since, $AG \cap EO = F$ Hence, $F$ is the circumcenter of $\Delta AMD$ and Since, $F$ lies on $C$ Hence, Proved!!