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Question

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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