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If Two Angles of a Triangle Are Unequal Then the Side Opposite to the Greater Angle Is Greater Than the Side Opposite to Smaller Angle

P Q R S is a ...

Question

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

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Solution

Let PQRS be the quadrilateral in which diagonals PR and QS intersect at O.

The sum of any two sides of a triangle is greater than the third side.
So, in $∆ P O Q, ∆ P O S, ∆ R O S and ∆ Q O R$ , we have
OP + OQ > PQ .....(i)
OP + OS > PS .....(ii)
OS + OR > RS .....(iii)
OR + OQ > QR .....(iv)
Adding (i), (ii), (iii) and (iv), we get
2(OP + OQ + OS + OR) > PQ + PS + RS + QR
⇒ 2(OP + OR + OQ + OS ) > PQ + PS + RS + QR
⇒ 2(PR + QS ) > PQ + PS + RS + QR
⇒ PQ + PS + RS + QR < 2(PR + QS)

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