$S=\odot R \cap \odot Q,T=SA\cap \odot R,DT$ passes $R$ $\angle TBA =\angle TCA=90^\circ -\angle A$ $\frac{Sin\angle BTA}{AB}=\frac{Sin\angle TBA}{AT}=\frac{Sin\angle TCA}{AT}=\frac{Sin\angle CTA}{AC}$ $\frac{BS}{SC}=\frac{Sin\angle BTA}{Sin\angle CTA}=\frac{AB}{AC}$ With $\angle BSC =\angle BDC=360^\circ-3\angle A$ we can now determine position of $S$ $\angle APO=180^\circ-2\angle AOP=2\angle C-2\angle B$(WLOG $AB>AC$) need to proof $\angle ASO=\angle C-\angle B$ $E$ is on ray $OS$ makes $OA^2=OS\cdot OE$ $OB^2=OC^2=OS\cdot OE\Rightarrow \triangle OBS \sim \triangle OEB,\triangle OCS \sim \triangle OEC$ $\Rightarrow \frac{EB}{EC}=\frac{BS}{CS}=\frac{AB}{AC}$ and $\angle BEC=\angle A$ $\Rightarrow AE\perp BC,\angle ASO=\angle OAE=\angle C-\angle B$