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If the diagonals PR and SQ of a quadrilateral PQRS meet at point O then Prove thatPQ+QR+RS+SP =2 (PR+QS) |
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Answer» Step-by-step explanation: given that , PQRS is a quadrilateral in which diagonal PR and QS intersect at a O . to prove - PQ +QR +RS+SP < 2 ( PR + QS ) proof - we know that SUM of any two SIDE of a triangle is greater than the third side . .'. in Δ PQO , PO+QO>PQ , .......................(i) in Δ SOP SO + PO >PS , .........................(ii) in Δ SOR SO + OR > RS ...........................(iii) in Δ QOR , QO + OR > QR ...........................(iv) on adding eqn. i , ii , iii & iv we get , PO+QO+SO+PO+SO+OR+QO+OR > PQ+PS+SR+QR also ⇒ 2 ( PO + QO + SO + OR ) > PQ+PS+SR + QR = 2( PR + QS ) > PQ+PS+RS + QR ( proved) .... |
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