/Bump because there isn't a solution

Let $TD$ intersect $AB$ at $F$ and let $DB$ intersect $CT$ at $E$. We have $\angle CTB = \angle TAB = 90 - \angle ATF = \angle BTF$. Thus, $TB$ is the angle bisector of $\angle FTC$ and $\angle ATB = 90$, so by a well-known theorem, $(F,C;B,A)=1$. Projecting those four points through $D$ onto line $CT$, we have $(F,C;B,A)=(T,C;E,CT_{\infty})=\frac{TE}{CE}=1$, so $TE=CE$ and thus $DB$ bisects $CT$.

Meh, Just Let $N=AB \cap TD$ and $E=BD \cap TC$ Now,we have $T,B,TD \cap k,A$ harmonic. Projecting through $D$ we get , $(P_{\infty},E;T,C)=-1$ and we are done.