# 5.  ABCD is quadrilateral. Is     $\small AB+BC+CD+DA< 2(AC+BD)$            ?

Let the intersection point of the two diagonals be O.

As we know that the sum of two sides of ANY triangle is always greater than the third side(Triangles Inequality Rule).

So,

In $\small \Delta AOB$ :

$\overline {AO}+\overline {OB}>\overline{AB}...........(1)$

In $\small \Delta BOC$ :

$\overline {BO}+\overline {OC}>\overline{BC}...........(2)$

In $\small \Delta COD$ :

$\overline {CO}+\overline {OD}>\overline{CD}...........(3)$

In $\small \Delta DOA$ :

$\overline {DO}+\overline {OA}>\overline{DA}...........(4)$

Now, Adding all four equations we, get

$\overline{AO}+\overline {OB}+\overline{BO}+\overline{OC}+\overline {CO}+\overline {OD}+\overline{DO}+\overline{OA}>\overline{AB}+\overline{BC}+\overline{CD}+\overline{DA}$

$2\left (\overline{AO}+\overline {OB}+\overline{OC}+\overline {OD} \right )>\overline{AB}+\overline{BC}+\overline{CD}+\overline{DA}$

$2\left (\left (\overline{AO}+\overline {OC} \right )+\left (\overline{CO}+\overline {OD} \right ) \right )>\overline{AB}+\overline{BC}+\overline{CD}+\overline{DA}$

$2\left (\overline{AC}+ \overline{BD} \right )>\overline{AB}+\overline{BC}+\overline{CD}+\overline{DA}$

which can also be expressed as

$\small AB+BC+CD+DA< 2(AC+BD)$

Hence this is true.

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