To: ABCD a parallelogram
Construction :Draw OX∥DC
In △BCD
O is the mid point of BD and OX∥DC
∴Byconverseofmidpointtheorem
X is the mid point of BC and OX=12DC
∴OX=12AB
Now in △ABC,
X is the mid point of BC and OX=12AB
∴Byconverseofmidpointtheorem
O is the mid point of AC and OX∥AB
As OX∥ABandOX∥DC
=AB∥DC
Now AB∥DCandAB=DC
∴ABCDisaparallelogram