Question

the diagonals of a quadrilateral PQRS intersect at O. prove that PQ2+SR2=PS2+QR2​

Answer 1

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  

Hence Proved.