Suppose $t_2$ is tangent to $\omega_1$ at point $P$, $\omega_1$ and $\omega_2$ are tangent at point $Q$. Suppose the tangent at $Q$ intersects $t_1$ and $t_2$ at $R$ and $S$ respectively. Since $\angle DRQ=\angle QSP$ and both $\triangle DRQ$ and $\triangle QSP$ are isosceles, $\angle RQD=\angle PQS$. This implies $D,P,Q$ are collinear. Now $\angle QPS=\angle QDR=\angle BCQ$, thus $B,C,P,Q$ are concyclic. By the intersecting chord theorem, $DB\cdot DC=DQ\cdot DP=DA^2$, which is equivalent to the circumcircle of $\triangle ABC$ is tangent to $t_1$.

Sorry if it is redundant, but anyway: Consider the inversion $I$ of pole$D$ and power $DA^2$. Then $\omega_1$ and $t_1$ are invariant under this inversion, while $\omega_2$ should transform to a line parallel to $t_1$ and tangent to $\omega_1$ hence in $t_2$. This means that $I(C)=B$ and hence $DA^2=DB\cdot{DC}$ and we are done.