IF PQRS is a quadrilateral, thens show that. PQ+QR+RS+PS is grater than 2( PR+QS)
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Step-by-step explanation:
We know that in any triangle the sum of any two sides is greater than the third side.
In △PQO,PQ+QO>PQ→(1)
In △SOP,SO+PO>PS→(2)
In △SOR,SO+OR>RS→(3)
In △QOR,QO+OR>QR→(4)
Adding equations (1) , (2) , (3) & (4) , we get
2(PO+OQ+SO+OR)>PQ+QR+RS+SP
PQ+QR+RS+SP<2(PQ+OR+SO+OQ)
PQ+QR+RS+SP<2(PR+SQ)
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