According to Geometry Problem (25), $ID$ passes through the midpoint of the A-altitude. Thus, using the extraversion of IMO ShortList 2002, geometry problem 7 on the A-excircle $\Gamma_a,$ we deduce that $\odot(SBC)$ is tangent to $\Gamma_a$ at $S.$ Let $SB,SC$ cut $\Gamma_a$ again at $B,'C'.$ Since $S$ is the exsimilicenter of $\Gamma_a \sim \odot(SBC),$ we have $BC \parallel B'C'$ $\Longrightarrow$ arcs $DB'$ and $DC'$ of $\Gamma_a$ are equal $\Longrightarrow$ $SD$ bisects $\angle BSC.$

Let $E,F$ be the other points of tangency of the ex-circle. Let $S'$ be the intersection of the circumcircles of $\triangle ICF$ and $\triangle IBE$. Then $E,F$ are reflections in $AI$, so $\angle IFC=\angle IEB$, which means $S'I$ bisects $\angle BS'C$. Also $\angle IS'F = C/2$, $\angle IS'E = B/2$, so $\angle FS'E = 90-A/2$, which means that $S'$ lies on the excircle also. But $\angle DS'F=\angle DEF = C/2$ as well, so $I,D,S'$ are collinear, which makes $S=S'$.

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See PP5 from here.

Let $W$ be the center of the ex-circle opposite $A$ and $L$ the midpoint of line $DS$. Let the incircle touch $BC$ at $E$ .We can easily prove that $\Delta EID \sim \Delta LDW \Rightarrow DL.DI =EI.DW$ Writing the formulas for A-ex-radius as $DW=\sqrt_ {\frac{s(s-b)(s-c)}{(s-a)}}$ and inradius as $\sqrt_ {\frac{(s-a)(s-b)(s-c)}{s}}$ gives $DL.DI=(s-b)(s-c)$ .But $DC.DB= (s-b)(s-c) \Rightarrow IBLC$ is cyclic..Let $R$ be the midpoint of $ID \Rightarrow RBSC$ is cyclic. Let the foot of perpendicular from $R$ on $BC$ be $J$ .Easy observation yields $JE=JD \Rightarrow JB=JC.\Rightarrow RB=RC \Rightarrow \angle RSB=\angle CBR=\angle RCB=\angle CSR$ as desired

An easy extension. Let $w=C(O,r)$ be a circle what is tangent to the line $d$ in the point $A\in d$ . Let $\{B,D,A,C\}\subset d$ in this order so that $BD=AC$ . Denote the point $E$ for which $ED\perp AB$ and $EB\perp OB$ . Define $F\in AE\cap w$ . Prove that the ray $[FA$ is the bisector of the angle $\widehat {BFC}$ . Proof. Denote the diameter $[AL]$ of $w$ and the midpoint $N$ , $M$ of the segments $[AE]$ , $[BC]$ respectively. Therefore, $\left\{\begin{array}{ccccc} \triangle BDE\sim\triangle OAB & \implies & \frac {BD}{OA}=\frac {DE}{AB} & \implies & AB\cdot AC=OA\cdot DE\\\\ \triangle EDA\sim\triangle AFL & \implies & \frac {ED}{AF}=\frac {EA}{AL} & \implies & OA\cdot DE=AN\cdot AF\end{array}\right|$ $\implies$ $AB\cdot AC=AN\cdot AF\implies$ $BNCF$ is cyclically. Since $NM$ is the bisector of $[BC]$ obtain that $NB=NC$ $\implies$ $[FA$ is the bisector of the angle $\widehat {BFC}$ .

Another solution:Let $Ia$ be the excenter and let $M$ be the midpoint of $ID$.Now,since it is well known that $BM=CM$ it is enough to prove that $BSCM$ is a cyclic or $DM*DS=DB*DC$,Now,since $BD*CD=r*Ra$(If $E$ the point where the incircle touches $BC$ we have that $BIE$ and $BIaD$ are similar) ,since $ID=2*DM$,let N be on $IaD$ such that $ND=r$ and let $K$ be on $Γa$ such that $DK$ is the diameter,it is enough to prove that $INSK$ is cyclic,but this is easy since $<INK=<ISK=90$,so we are finished.

My solution: Let $ I_a $ be $ A- $ excenter of $ \triangle ABC $ . Let $ M $ be the midpoint of $ DS $ and $ T=I_aM \cap BC $ . Easy to see $ TS $ is tangent to $ (I_a) $ at $ S $ . Since $ \angle IBI_a=\angle ICI_a=\angle IMI_a=90^{\circ} $ , so we get $ I, I_a, B, C, M $ lie on a circle with diameter $ II_a $ . Since $ TB \cdot TC=TM \cdot TI_a=TD^2=TS^2 $ , so $ \triangle TSB \sim \triangle \Longrightarrow TCS \angle CSD=\angle SCT-\angle SDT=\angle TSB-\angle TSD=\angle DSB $ . Q.E.D