Abcd is a Quadrilateral and AC is one of its diagonals . Prove that AB+BC+CD+DA>AC+BD
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Step-by-step explanation:
In ADC
AD+DC>AC (SUM OF TWO SIDES OF A TRIANGLE IS GREATER THAN THE THIRD SIDE) -(1)
SIMILARLY,
IN DBC
DC+BC>BD-(2)
IN ABC
AB+BC>AC-(3)
IN ADB
AD+AB>BD-(4)
adding, (1)+(2)+(3)+(4)
AD+DC+DC+BC+AB+BC+AD+AB>AC+DB+AC+DB
=2(AB+BC+CD+DA)>2(AC+BD)
=AB+BC+CD+DA>AC+BD
Hence proved
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