Given: PR∥BS, P is mid point of AB and Q is mid point of DR
2AD=AB
2AD=AP+PB
2AD=2AP (P is mid point of AB)
AD=AP
or A is mid point of PD
Now, In △DPR,
A is mid point of PD and Q is mid point of QR. Thus, by converse of mid point theorem,
AQ∥PR
Since, PR∥BS
hence, AQ∥PR∥BS