Question

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ABCD is a quadrilateral whose diagonals AC and BD intersect at O.Prove that
(AB+BC+CD+DA)<2 (AC+BD)

Answer 1

since the sum of two sides of a triangle is greater than third side
ao+ob>ab
Ob +oc>BC
od +oc>DC
od +ao>ad
adding all eqn
ab +BC+cd+ad<2{(do+Ob)+(oc+oa)}
ab +BC+cd+ad<2(AC+bd)

Answer 2

ao+ob>AB 1
OD+OC>DC 2

OD +AO>AD 3
ADDIND THE EQUATION
AB+ +CD+AD< 2 ((OD+OD)+(OC OA)
AB+BC +CD +AD <2(AC+BD)
IS. IT OK