let $XC\cap (ABC)=Y$. by applying Pascal's theorem in points$A, S_x, C, B, T_x, Y$, $CY\cap S_xA=X$, $AB\cap YT_x=F_x'$, $T_xS_x\cap BC=D_x$. $X, F_x', D_x$ are collinear. Thus, $F_x=F_x'$. Then, $Y, T_x, F_x$ are collinear. (1) Since $D_xX^2=D_xB\cdot D_xC=D_xT_x\cdot D_xS_x$, $\angle D_xXT_x=\angle XS_xT_x=\angle ACT_x$. Thus, $(E_x, X, T_x, C)$. (2) let $l$ be the tangent line of $(E_xT_xF_x)$ passing through $T_x$. let $K=l \cap AX, L=l\cap BC$. It suffices to show that $l$ is also tangent to $(ABC)$. $\angle F_xE_xT_x=\angle F_xT_xL=\angle YT_xK$ by (1) by (2) $\angle F_xE_xT_x=\angle YCT_x$. Since $\angle YCT_x= \angle YT_xK$, $l$ is tangent to $(ABC)$ which ends the problem.