In the figure, prove that:
$CD + DA + AB + BC > 2AC$

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To do:

We have to prove that $CD + DA + AB + BC > 2AC$.

Solution:

In the given figure,

$ABCD$ is a quadrilateral and $AC$ is joined.

In $\triangle ABC$,

$AB + BC > AC$.........…(i)             (Sum of two sides of a triangle is greater than its third side)

Similarly,

In $\triangle ADC$,

$CD + DA > AC$......…(ii)

Adding (i) and (ii), we get,

$CD + DA + AB + BC > AC + AC$

$CD + DA + AB + BC > 2AC$.

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Updated on: 10-Oct-2022

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