Denote the vertices of $T$ like the figure. We have $F$ is the incenter of $\Delta ABC$, $E$ is the $B$-excenter of $\Delta ABX$ and $D$ is the incenter of $\Delta AXC$ a) $\widehat{EDF}=\widehat{CDX}=90^o+\dfrac{\widehat{XAC}}{2}$ $\widehat{EAF}=\widehat{EAB}-\widehat{BAF}=90^o+\dfrac{\widehat{XAB}}{2}-\dfrac{\widehat{A}}{2}=90^o-\dfrac{\widehat{XAC}}{2}$ Therefore $D,E,F,A$ are concyclic. b) Let $G$ be the second intersection of $(DEF)$ and $AB$. We have $\widehat{GDF}=\widehat{GAF}=\dfrac{\widehat{A}}{2}$ $\widehat{DFE}=180^o-\widehat{BFC}=90^o-\dfrac{\widehat{A}}{2}$ $\Rightarrow \widehat{GDF}+\widehat{DFE}=90^o$ or $DG \perp EF$ Denote $H$ by the orthocenter of $\Delta DEF$. Since $G \in (DEF)$ and $DG \perp EF$, we have $G$ is the symmetric point of $H$ about $EF$ Also, $G \in AB$ and $EF$ is the bisector of $\widehat{ABC}$, hence $H \in BC$