Question

Given that $ABCD$ is a quadrilateral. Is $AB+BC+CD+DA<2(AC+BD)?$

Answer 1

As we know that the sum of lengths of any two sides in a triangle should be greater than the length of third side-

Therefore,

In $△AMB,AB<MA+MB$ …….(i)

In $△BMC,BC<MB+MC$ …….(ii)

In $△CMD,CD<MD+MC$ ..….(iii)

In $△AMD,DA<MD+MA$ …….(iv)

On adding equation (i), (ii), (iii) and (iv), we get

$AB+BC+CD+DA<2MA+2MB+2MC+2MD$

$AB+BC+CD+DA<2\left[\right(AM+MC)+(DM+MB\left)\right]$

Since, $AC=AM+MCandBD=BM+MD$

Hence,

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

It is proved.