ABCD us a quadrilateral. Prove that

AB+BC+CD+DA>AC+BD

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Similar questions

A B C D

A B + B C + C D + D A > A C + B D

$A B C D$ is a quadrilateral
Prove that $(A B + B C + C D + D A) > (A C + B D)$

In the adjoining figure, ABCD is a quadrilateral and AC is one of its diagonals. Prove that

(i) AB + BC + CD + DA > 2AC

(ii) AB + BC + CD > DA

(iii) AB + BC + CD + DA > AC + BD.

Let ABCD be a quadrilateral with diagonals AC and BD. Prove the following statements (Compare these with the previous problem);

(a) AB + BC + CD > AD;

(b) AB + BC + CD + DA > 2AC;

(d) AB + BC + CD + DA > AC + BD.

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