Question

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

Answer 1



$ABCD$ is a quadrilateral.

Then, $AC$ and $BD$ are the diagonals.

Now, the both diagonals are divided by a quadrilateral in four triangles.

We know that, the sum of two sides of a triangle is greater then the third side. Therefore,

In $△DOC,DO+OC>DC$...(1)

Similarly,

$△BOC,BO+OC>BC$...(2)

$△BOA,BO+OA>AB$...(3)

$△AOD,DO+OA>AD$...(4)

Adding, (1),(2),(3) and (4), we get,

$\Rightarrow 2(AO+OB+OC+OD)>AB+BC+CD+DA$

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

Hence, proved.