A) \[\left( AB+BC+CD+DA \right)<\left( AC+BD \right)\]
B) \[\left( AB+BC+CD+DA \right)>2\left( AC+BD \right)\]
C) \[\left( AB+BC+CD+DA \right)>\left( AC+BD \right)\]
D) \[AB+BC+CD+DA=2\left( AC+BD \right)\]
Correct Answer: C
Solution :
(c): In \[\Delta ABC,\Delta ACD,\Delta BCD\] and \[\Delta ABD\] \[AB+BC>AC\] \[CD+DA>AC\] \[BC+CD>BD\] \[DA+AB>BD\] Adding above inequalities \[2\left( AB+BC+CD+DA \right)>2\left( AC+BD \right)\] \[\Rightarrow \]\[AB+BC+CD+DA>\left( AC+BD \right)\]
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