PQRS IS A QUAD. IN WHICH DIAGONALS PR AND QS INTERSECT AT O.PROVE THAT PQ+QR+RS+SP>2(PR+QS).

Hint:

Use the triangular inequality,
a+b>c
where a,b,c represent three sides of any triangle.

To prove that PQ + QR + RS + SP > 2(PR + QS) when the diagonals PR and QS of quadrilateral PQRS intersect at point O, we can use the Triangle Inequality Theorem and the properties of a quadrilateral.

Here's how we can approach the proof:

1. Let's label the lengths of the sides of quadrilateral PQRS as follows:
PQ = a
QR = b
RS = c
SP = d

2. We know that the diagonals PR and QS intersect at point O. This implies that triangles POR and QOS are formed.

3. Using the Triangle Inequality Theorem, we can state that for any triangle, the sum of the lengths of any two sides is always greater than the length of the third side. Applying this to triangles POR and QOS, we have:
PR + RO > PO,
PR + PO > OR,
QS + SO > QO,
QS + QO > OS.

4. Adding the above four inequalities, we get:
PR + RO + PR + PO + QS + SO + QS + QO > PO + OR + OS.
Simplifying this expression, we have:
2(PR + QS) + (RO + PO + SO + QO) > PO + OR + OS.

5. Now, let's focus on the right-hand side of the inequality. By applying the Triangle Inequality Theorem again, we know that RO + PO > RP and SO + QO > SQ. Thus, we can replace them in the inequality:
2(PR + QS) + (RP + SQ) > PO + OR + OS.
2(PR + QS) + (RP + SQ) > PQ + QR + RS + SP.

6. Finally, by substituting the lengths of the sides of the quadrilateral into the inequality, we get:
2(a + c) + (b + d) > a + b + c + d.
2a + 2c + b + d > a + b + c + d.
a + c > 0.

Since a + c is always greater than 0, we can conclude that PQ + QR + RS + SP > 2(PR + QS).

This completes the proof.