To prove the given statement, we will use the concept of the properties of diagonals in a quadrilateral.

Given: PQRS is a quadrilateral and diagonals PR and QS intersect at O.

To prove:
(i) PQ + QR + RS + SP > PR + QS
(ii) PQ + QR + RS + SP > PR + QS

Proof:
(i) To prove PQ + QR + RS + SP > PR + QS, we will use the property that in a quadrilateral, the sum of the lengths of any two opposite sides is always greater than the sum of the lengths of the diagonals.

We can divide the quadrilateral PQRS into two triangles, namely triangle POR and triangle QOS, by drawing the diagonals PR and QS.

- Triangle POR: In triangle POR, PQ and QR are two sides, and PR is the diagonal.
- Triangle QOS: In triangle QOS, RS and SP are two sides, and QS is the diagonal.

Using the property of triangles, we can write the inequalities:
PQ + QR > PR
RS + SP > QS

Adding the above two inequalities, we get:
PQ + QR + RS + SP > PR + QS

Therefore, PQ + QR + RS + SP is greater than PR + QS.

(ii) To prove that PQ + QR + RS + SP > PR + QS, we will use the concept of the triangle inequality.

In triangle POR, we have:
PQ + QR > PR

In triangle QOS, we have:
RS + SP > QS

Adding the above two inequalities, we get:
PQ + QR + RS + SP > PR + QS

Therefore, PQ + QR + RS + SP is greater than PR + QS.

Hence, we have proved that in the quadrilateral PQRS, PQ + QR + RS + SP is greater than PR + QS.