In a quadrilateral ABCD, show that


In a quadrilateral ABCD, show that (AB + BC + CD + DA) > (AC + BD).


Given: Quadrilateral ABCD

To prove: (AB + BC + CD + DA) > (AC + BD)


In $\Delta A B C$,

$A B+B C>A C \quad \ldots$ (i)

In $\triangle C A D$,

$C D+A D>A C \quad \ldots$ (ii)

In $\Delta B A D$,

$A B+A D>B D \quad \ldots$ (iii)

In $\Delta B C D$

$B C+C D>B D \quad \ldots$ (iv)

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

2(AB + BC + CD + DA) < 2( AC + BD)

Hence, (AB + BC + CD + DA) < (AC + BD).

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