Construction : Draw a line PR parallel to CD , R lies on AD.
From corollary to mid point theoram in triangle ADC , as PR|| CD, P being the mid point , R is the mid point of AD.
As CD||AB, we have PR|| CD..Consider the triangle ABC, we have R as mid point of AD and similar as above it meets BD at midpoint i.e Q. Thus PQ||AB.
Note: Above solution assumes that CD is the larger side.