Question

ABCD is a quadrilateral. show that AB+BC+CD+DA is greater than AC+BD​

Answer 1

Taking diagonals AC and BD

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Answer 2

In ΔABC,

AB+BC>AC (Inequalities in triangles) ---> (1)

Similarly, in ΔBCD,

BC+CD>BD  (Inequalities in triangles) ---> (2)

Similarly, in ΔCDA,

CD+DA>AC  (Inequalities in triangles) ---> (3)

Similarly, in ΔDAB,

DA+AB>BD  (Inequalities in triangles) ---> (4)

Add equations (1), (2), (3), (4),

AB+BC+BC+CD+CD+DA+DA+AB>AC+BD+AC+BD

=2AB+2BC+2CD+2DA>2AC+2BD

=2(AB+BC+CD+DA)>2(AC+BD)

=AB+BC+CD+DA>AC+BD

 Hence Proved           :)