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
 A: 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 :
$$\frac{R_b}{R_c}=\frac{AB}{2\sin\angle{ADB}}\cdot\frac{2\sin\angle{ADC}}{AC}=\frac{AB}{AC}\cdot\frac{\sin\angle{ADC}}{\sin\angle{ADB}}=\frac{AB}{AC}\cdot\frac{\sin(180^\circ-\angle{ADB})}{\sin\angle{ADB}}=\cdots$$
A: Applying sine rule in $\triangle ABD$ as follows $$\frac{\sin\angle ABD}{AD}=\frac{\sin\angle ADB}{AB} $$ $$\implies \color{red}{\sin \angle ABD=\frac{AD}{AB}\sin\angle ADB}$$ Now, the circumscribed radius $R_b$ of $\triangle ABD$ is given as $$R_b=\frac{\text{side of triangle }}{2\times\sin \text{(opposite angle)}}$$$$=\frac{AD}{2\sin \angle ABD}=\frac{AD}{2\times \frac{AD}{AB}\sin\angle ADB }$$$$R_b=\frac{AB}{2\sin \angle ADB}\tag 1$$
Similarly, applying sine rule in $\triangle ACD$ as follows 
$$\frac{\sin\angle ACD}{AD}=\frac{\sin\angle ADC}{AC} $$ $$\implies \color{red}{\sin \angle ACD=\frac{AD}{AC}\sin\angle ADC}$$ Now, the circumscribed radius $R_c$ of $\triangle ACD$ is given as $$R_c=\frac{AD}{2\sin \angle ACD}=\frac{AD}{2\times \frac{AD}{AC}\sin\angle ADC} $$  $$R_c=\frac{AC}{2\sin \angle ADC}$$ But, we have $$\angle ADB+\angle ADC=180^\circ\iff \angle ADC=180^\circ-\angle ADB$$ Now, setting the value of $\angle ADC$, we get   $$ R_c=\frac{AC}{2\sin(180^\circ- \angle ADB)}$$ $$R_c=\frac{AC}{2\sin \angle ADB}\tag 2$$
Now, dividing (1) by (2), we get $$\frac{R_b}{R_c}=\frac{\frac{AB}{2\sin \angle ADB}}{\frac{AC}{2\sin \angle ADB}}=\frac{AB}{AC}$$ 
Hence, proved that 
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\frac{R_b}{R_c}=\frac{AB}{AC}}}$$
A: For short:
$$ \frac{R_b}{R_c}=\frac{\frac{AB}{\sin\widehat{ADB}}}{\frac{AC}{\sin\widehat{ADC}}}=\frac{AB}{AC}$$
by the extended sine theorem, since $\widehat{ADB}$ and $\widehat{ADC}$ are supplementary angles, so they have the same sine.
