Since $S$ is tangent point, so $\angle{RPS}=\angle{PBS}$. Also since $AB$ is tangent to the circle, so $\angle{PRS}=\angle{BPS}$. Therefore $\triangle{BPS} \sim \triangle{PRS}$. Similarly, we have $\angle{BSP}=\angle{ASP}$. Similarly we have: $\angle{CSQ}=\angle{ASQ}$. Thus: \[ \frac{BP}{BS}=\frac{AP}{AS}=\frac{AQ}{AS}=\frac{CQ}{CS} \] which implies $\frac{BP}{CQ}=\frac{BS}{CS}$. Also, we have: \[ \frac{PT}{QT}=\frac{\sin{\angle{BAS}}}{\sin{\angle{CAS}}}=\frac{BS}{CS} \] togather with $\angle{BPT}=\angle{CQT}$, we conclude that $\triangle{BPT} \sim \triangle{CQT}$. Thus $\angle{BTP}=\angle{CTQ}$.

We can construct the common targent $ Sx$ of two circle to proof We know that PQ pass through $ I$( $ I$ is the incenter of incircle $ ABC$). Hence, $ SBPI$ and $ SCQI$ are cyclic quadrilater, we get the result

Dear Mathlinkers, Circle gamma is the A-mixtilinear incircle of ABC. What can we say with the A-mixtilinear excircle of ABC? Sincerely Jean-Louis

Some other solutions: http://www.mathlinks.ro/viewtopic.php?t=140464

Dear Mathlinkers, you can see my synthetic proof on my website http://perso.orange.fr/jl.ayme vol. 4 A new mixtilinear incircle adventure p. 56-58 Sincerely Jean-Louis

jayme wrote: Dear Mathlinkers, you can see my synthetic proof on my website http://perso.orange.fr/jl.ayme vol. 4 A new mixtilinear incircle adventure p. 56-58 Sincerely Jean-Louis Few people understand the language. Can you post them here in English?

Trivial from $S,I,P,B$ and $S,T,Q,C$ cyclic and $ST$ is symmedian in $SPQ$