This is quite easy let $TM \cap QR=N'$ now by simple similiarty prove $N'$ is the midpoint of $QR$ so $N' \equiv N$ and we are done $\blacksquare$.

$AD\cdot CD=PD\cdot DR\implies\frac{AD}{PD}=\frac{DR}{CD}$ and $\angle ADP=\angle CDR$ which implies $\Delta ADP\sim \Delta CDR\implies AP\parallel CR$. Let $X=AP\cap BC$ and $Y=CR\cap PQ$. So, $XPYC$ is a parallelogram. $\Delta ABX\sim\Delta CDX$ and $XM$ is the median of $\Delta CDX$ which means $TX$ is the median of $\Delta ABX$. Thus, $X,T,M$ are colinear. Similarly, $M,N,Y$ are colinear. But $XPYC$ is parallelogram which means $X,M,Y$ are colinear. Hence, proved.