prove that in a quadrilateral the sum of all the sides is greater than the sum of its diagonals
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The triangle inequality assures us that AB+BC>ACAB+BC>AC, BC+CD>BDBC+CD>BD, CD+AD>ACCD+AD>ACand AD+AB>BDAD+AB>BD. Adding all these gives 2AB+2BC+2CD+2AD>2AC+2BD2AB+2BC+2CD+2AD>2AC+2BD. Dividing by 2, AB+BC+CD+AD>AC+BDAB+BC+CD+AD>AC+BDas
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