Quote: $E$ is the midpoint of $AC$ It should be that "$E$ is the midpoint of $AX$". Suppose that the line $AD$ cut $TR$ at point $X$. we want to show that $DX$ = $DC$, hence $X$ is the incenter of $\triangle ABC$. First note that $\angle RDX=\angle SDA=\angle SEA$ and $\angle RXD=\angle XDE =\angle ADE =\angle ASE$, and hence we have $\triangle SAE$ similiar to $\triangle XRD$. Now we can compute $DX$, on the condition that $\frac{SE}{TE}=\frac{SD}{RD}$: \[ DX^2 = \frac{RD^2*SE^2}{AE^2}=\frac{RD^2*SE*SE}{SE*TE}=\frac{RD^2*SE}{TE}=\frac{RD^2*SD}{RD} = SD*RD = DC^2 \] and we are done. Note: I use the results that $CD$ is tangent to ciccumcircle of $\triangle SCD$ and $EA$ is tangent to circumcircle of $\triangle SAT$.

Could You Post The Figure ( Shape ) Because Didn't Manage To Complete it ?

thx to Lei Lei, I can give this problem to my stus

Notice that X is symmetric to B, we can also pick up the incenters of triangle ACB and ACX.

S should be on arc AC, not AX.. well, it doesn't matter much, but when you say AX, it looks like ABX and the figure looks impossible afterwards. By the way, what happened to CMO 2011 this year? It looks awfully simple and reused - hopefully the team selection tests are up to their usual qualifications/beauty.

It's famous Lemma for THEBAULT's THEOREM see it here : http://jl.ayme.pagesperso-orange.fr/Docs/Sawayama%20or%20Thebault's%20Theorem.pdf By Jean-Louis AYME

Denote by $K$ the intersection of $AD$ and $BC$. It's easy to see that $AKRS$ is cyclic because $$\angle AKR + \angle ASR = \frac{1}{2}\angle BAC + \angle ABC + \angle ACB + \frac{1}{2} \angle BAC = 180^{\circ}$$Let $AD$ intersect $TR$ at $I$. Now some angle chasing gives that $\angle ADE = \angle DIR = \angle ISR$ and of course $\angle CAD = \angle DCK$ which gives that $DI^{2} = DR \cdot DS = DK\cdot AD = CD^{2}$. It is well-known that $I$ is the incentre because $BD = CD = DI$ and $DI$ is the bisector of $\angle BAC$.

Cute problem, Firstly , we have that $\triangle STR$ and $\triangle SED$ are homothetic and thus $( STR)$ is tangent to $(ABC)$ at $S$ .Now, we have that $\angle RTX=\angle (TX,DE)=\frac{1}{2} (\widehat AE + \widehat DX)=\frac{1}{2} (\widehat EX+ \widehat XD)=\frac{1}{2}(\widehat DE)=\angle ESD$. Thus $AT$ is tangent to $(STR)$ and similarly $BC$ is tangent to $(STR)$ . Thus,$(STR)$ is a curvilinear incircle ,which gives the desired.