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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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