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]
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Draw PA and RB so that they meet PQ and SR (By const.)
In quardilateral PARB
1/2 angle P = 1/2 angle R
 angle BPR = angle BRA (PA &RB are angle bisectors)
so PBllAR
therefore PBRS  is a llgm with opposite side equal and opposite side equal
so PAllRB (opposite sides of a llgm are equal )

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