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Question

$$PQRS$$ is a quadrilateral in which the diagonals intersect each other at O. Prove that $$PQ + QR + RS + SP < 2(PR+QS)$$


Solution

We know that in any triangle the sum of any two sides is greater than the third side.
In $$\bigtriangleup PQO , PQ+QO>PQ\rightarrow (1)$$
In $$\bigtriangleup SOP , SO+PO>PS\rightarrow (2)$$
In $$\bigtriangleup SOR , SO+OR>RS\rightarrow (3)$$
In $$\bigtriangleup QOR , QO+OR>QR\rightarrow (4)$$
Adding equations (1) , (2) , (3) & (4) , we get
$$2(PO+OQ+SO+OR)>PQ+QR+RS+SP$$
$$PQ+QR+RS+SP<2(PQ+OR+SO+OQ)$$
$$PQ+QR+RS+SP<2(PR+SQ)$$

969064_1029714_ans_51ca1c7e54c84b348ac17094739dc4ad.png

Mathematics

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