By either Ratio lemma or homothety it is well known that if you let $I$ incenter of $\triangle ABC$, $D,E,F$ touch points of the incircle $\omega$ with $BC,AC,AB$ respectively, $M$ midpoint of $BC$ then you have that $AM,ID,EF$ are concurrent (call $R$ the point of concurrency). Now draw the second tangent from $M$ to $\omega$ and let it hit it at $D_1$, let $ID \cap \omega=A'$. Taking the dual of the first mentioned concurrency we get that $DD_1,EF, A\infty_{BC}$ are concurrent at $J$. Now $-1=(P, Q; \infty_{BC}, M) \overset{A}{=} (X, Z; A\infty_{BC} \cap XZ, AM \cap XZ)$, denote $\mathcal P_{\alpha}$ as the polar of $\alpha$ w.r.t. $\omega$, we have $J \in \mathcal P_{AM \cap XZ}$ by La'Hire but also by this cross ratio we get $A\infty_{BC} \cap XZ \in \mathcal P_{AM \cap XZ}$ so we get that $\mathcal P_{AM \cap XZ}=A\infty_{BC}$ which means $AM \cap XZ=R$, apply the same for $YT$ we get $XZ \cap YT=R$ which is fixed,thus we are done .

See https://artofproblemsolving.com/community/c6h616635 for a generalization.