Let ABCD be a quadrilateral in which the diagonals intersect at O perpendicularly. Prove that AB + BC + CD + DA > AC + BD.

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Solution

We know that in any triangle, the sum of two sides is always greater than the third side.

From the figure, we get:

AB + BC > AC

BC + CD > BD

CD + AD > AC

AD + AB > BD

Adding all the inequalities, we get:

2 (AB + BC + CD + AD) > 2 (AC + BD)

AB + BC + CD + AD > AC + BD

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Similar questions

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.

A B^{2} + C D^{2} = B C^{2} + D A^{2}

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.

ABCD us a quadrilateral. Prove that

AB+BC+CD+DA>AC+BD

Diagonal AC and BD of quadrilateral ABCD intersects each other at point O.

Prove that AB + BC + CD + AD < 2 (AC + BD).

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