Let $ J,K$ be the incenters of triangles $ ABC, DBC, L,M$ be the tangencies of $ \Gamma$ and $ BD, AC, Q$ be the midpoint of arc $ BC$ that contains $ T$ Applying Lyness's theorem for quadrilateral $ ABCD$ we have $ L,M, J,K$ are collinear. $ \angle BJC=\angle BKC=90^o-\angle BQC$ therefore $ Q$ is the circumcenter of $ BJKC$. Denote $ P=BC\cap TQ$ It easy to check that $ TQ$ is the externally bisector of angle $ BTC$ then $ \frac{PB}{PC}=\frac{BT}{CT}$ On the other hand, denote $ P'=BC\cap LM$ then applying Menelaus's theorem we obtain: $ \frac{P'B}{P'C}=\frac{BL}{CM}$ This result is well-known: $ \frac{BL}{CM}=\frac{BT}{CT}$ thus $ P'\equiv P$ or $ BC, LM, QT$ concur at $ P$. Hence $ PT.PQ=PB.PC=PL.PK$ Then $ T$ lies on the polar of $ P$ wrt $ (BLKC)$ we get $ T, R=BJ\cap CK, S=CJ\cap BK$ are collinear.

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Nice dear livetolove212 actually, there are plenty of results of this nice figure : X is the incenter of ABC Y is incenter of DBC XPSTB are concyclic QYSTC are concyclic TS pass through the midpoint of bigger arc BC

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You can use the two times the result proved on the following link and you are done. http://www.mathlinks.ro/viewtopic.php?search_id=1338641321&t=6086

Let P be the tangencies of T and AC , and Q be the tangencies of T and DB. Let M is Incenter of trangle ABC and n is Incenter of trangle DBC. Use Theorem of Viktor Tebo. P , Q , M , N lying in the same straight line. Trivial that RMQB is concyclic and RNPC is concyclic. \angle EQP= \angle QPE= \angle MSB , so SMQB and CPNS are concyclic , so RBQMS are concyclic RCNPS are concyclic. If line TS\cap circle of RCNPS = G. Then TS*TG=TN*TC and easy to see that CNMB are concyclic . So TN*TC=TM*TB , so TS*TG=TM*TB , so SMBG are concyclic , so G=R , so T, S , R are collinear

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Dear Giorgieri and Mathlinkers, this nice problem can be an interesting application of the article http://perso.orange.fr/jl.ayme Un remarquable résultat de Vladimir Protassov p. 7-10 Sincerely Jean-Louis