In a quadrilateral PQRS, the diagonals PR and QS intersect each other at O. Show that
(i) PQ + QR + RS + SP > PR + QS.
(ii) PQ + QR + RS + SP < 2 (PR + QS). ![](https://gs-post-images.s3.amazonaws.com/user_files/63093/14901/images/6ca65614fb093f4ad7324b896373192a.png)
We know that in a triangle the sum of two sides of a triangle is greater than the third side. Therefore in ΔPSR, we have
PS + SR > PR ...... (i)
In ΔQSR, QR + RS > QS ...... (ii)
In ΔPQS, PQ + SP > QS ...... (iii)
In ΔPQR, PQ + QR > PR ..... (iv)
Adding (i), (ii), (iii) and (iv), we get
2 (SP + PQ + QR + RS) > 2 (QS + PR) PQ + QR + RS + SP > PR + QS
Similarly in ΔPQO, ΔQRO, ΔRSO and ΔSPO,
we have
QO + PO > PQ ...... (v)
QO + RO > QR ...... (vi)
RO + SO > RS ...... (vii)
SO + PO > SP ...... (viii)
Adding (v), (vi), (vii) and (viii),
we get
2 (QO + SO + PO + OR) > PQ + QR + RS +SP 2 (QS + PR) > PQ + QR + RS + SP
or PQ + QR + RS + SP < 2 (QS + PR).