Prove that in a quadrilateral the sum of all the sides is greater than the sum of its diagonals.

To do:

We have to prove that in a quadrilateral the sum of all the sides is greater than the sum of its diagonals.

Solution:

Let in a quadrilateral $ABCD$, $AC$ and $BD$ are its diagonals.

In $\triangle ABC$,

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

Similarly,

In $\triangle ADC$,

$DA + CD > AC$........…(ii)

In $\triangle ABD$,

$AB + DA > BD$..........…(iii)

In $\triangle BCD$,

$BC + CD > BD$.........…(iv)

Adding equations (i), (ii), (iii) and (iv), we get,

$2(AB + BC + CD + DA) > 2AC + 2BD$

$2(AB + BC + CD + DA) > 2(AC + BD)$

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

Hence proved.

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Updated on: 10-Oct-2022

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