Proving a ratio that has a relation with the Perpendicular bisectors and circumcircle $ABC$ is a triangle, $D$ is a point on the side $BC$ of  $\triangle ABC$, $R_b$ is circumradius of $\triangle ABD$ , and $R_c$ is the  circumradius of $\triangle ACD$. 
Prove that 
$$ {Rb\over Rc} ={AB\over AC}$$
Thanks for your help

HINT : 
By the law of sines, we have$$R_b=\frac{AB}{2\sin\angle{ADB}}\quad\text{and}\quad R_c=\frac{AC}{2\sin\angle{ADC}}$$
with$$\angle{ADC}+\angle{ADB}=180^\circ.$$
Added :
$$\frac{R_b}{R_c}=\frac{AB}{2\sin\angle{ADB}}\cdot\frac{2\sin\angle{ADC}}{AC}=\frac{AB}{AC}\cdot\frac{\sin\angle{ADC}}{\sin\angle{ADB}}=\frac{AB}{AC}\cdot\frac{\sin(180^\circ-\angle{ADB})}{\sin\angle{ADB}}=\cdots$$
