Let ∠P = 2x degrees
.`. ∠R = 2x degrees (as opp angles of a parallelogram are equal)
.`. ∠APB = ∠ ARB = 2x/2 = x degrees (as PA and RB bisects ∠P and ∠ R respectively [given])
A line segment PQ is drawn through the points P and Q such that it bisects the angles ∠APB and ∠ARB.
Let PQ be a transversal.
Alternate angles ∠PRB and ∠APR are formed.
∠PRB = ∠ APR = x /2 degrees (as PQ bisects angles ∠APB and ∠ARB)
.`. Alternate angles are equal.
.`. PA || RB [Proved]