Let O be the point of intersection of the angular bisectors AP, and QB
We know that,
1) Angular bisectors intersect at 90∘.
2) Angular bisectors bisect each other, i.e., AO=OP and BO=OQ
Join QP, to form △POQ
Consider △AOB and △POQ,
AO=OP
∠AOB=∠POQ=90∘
BO=OQ
The two triangles are congruent. Hence the side AB and QP must be equal. Hence, parallel.
We know that, in parallelogram ABCD,
AB=CD and AB∥CD
Hence, CD=QP and CD∥QP. Proved.