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Question
Diagonal AC and BD of quadrilateral ABCD intersects each other at point O.
Prove that AB + BC + CD + AD < 2 (AC + BD).
Diagonal AC and BD of quadrilateral ABCD intersects each other at point O.
Prove that AB + BC + CD + AD < 2 (AC + BD).
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Solution
In a quadrilateral ABCD, with diagonals intersecting at O. AO+OB>AB(INEQUALITY)
OD+OA>AD
OD+OC>DC
OB+OC>BC
ADDING
AO + OB + OA + OD + OC + OB + OC > AB + AC + BC + CD
(AC+BD+AC+BD)>AB+BC+CD+AD
2(AC+BD)>AB+AC+BC+CD
HENCE YES THE STATEMENT IS TRUE
In a quadrilateral ABCD, with diagonals intersecting at O. AO+OB>AB(INEQUALITY)
OD+OA>AD
OD+OC>DC
OB+OC>BC
ADDING
AO + OB + OA + OD + OC + OB + OC > AB + AC + BC + CD
(AC+BD+AC+BD)>AB+BC+CD+AD
2(AC+BD)>AB+AC+BC+CD
HENCE YES THE STATEMENT IS TRUE
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