The bisectors of the angles of a Parallelogram enclose aa)Rhombusb)Squ...
We have ABCD, a parallelogram given below:

Therefore, we have AB || BC
Now, AD || BC and transversal AB intersects them at A and B respectively. Therefore,
Sum of consecutive interior angle is supplementary. That is;
∠A + ∠B = 180°

We have AR and BR as bisectors of ∠A and ∠B respectively.
∠RAB +∠RBA = 90° …… (i)
Now, in ΔABR, by angle sum property of a triangle, we get:
∠RAB + ∠RBA +∠ARB = 180°
From equation (i), we get:
90° + ∠ARB = 180°
∠ARB = 90°
Similarly, we can prove that ∠DPC = 90° .
Now, AB || DC and transversal ADintersects them at A and D respectively. Therefore,
Sum of consecutive interior angle is supplementary. That is;
∠A + ∠D = 180°

We have AR and DP as bisectors of ∠A and ∠D respectively.
∠DAR + ∠ADP = 90° …… (ii)
Now, in ΔADR, by angle sum property of a triangle, we get:
∠DAR + ∠ADP + ∠AQD = 180°
From equation (i), we get:
90° +∠AQD = 180°
∠AQD = 90°
We know that ∠AQD and ∠PQR are vertically opposite angles, thus,
∠PQR = 90°
Similarly, we can prove that ∠PSR = 90° .
Therefore, PQRS is a rectangle.
Hence, the correct choice is (d).