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Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD

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Given, ABCD is a quadrilateral

To Prove : AB + BC + CD + DA > AC + BD

Proof - Since we know that the sum of the two sides of a triangle is greater than the third side.

In \triangleABC, AB + BC > AC

In \triangleBCD, BC + CD > BD

In \triangleCAD, AD + CD > AC

In \triangleBAD, BA + AD > BD

Adding all the above equations,

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

\Rightarrow 2(AB + BC + CA + AD) > 2(AC + BD)

\Rightarrow AB + BC + CA + AD > AC + BD

Hence, proved.

Posted by

infoexpert24

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