Question

$ABCD$ is a quadrilateral
Prove that $(AB+BC+CD+DA)>(AC+BD)$

Answer 1

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

$ABCD$ is a quadrilateral and $AC$ and $BD$ are diagonals.

We know that sum of the two sides of $△$ is greater than the third side. Considering $△ABC,△CAD,$ $△BAD,$ and $△BCD,$ we have

$AB+BC>AC$

$CD+AD>AC$

$AB+AD>BD$

$BC+CD>BD$

Adding all the above equation

$2(AB+BC+CD+AD)>2(AC+BD)$

$\Rightarrow (AB+BC+CD+AD)>(AC+BD)$ (proved)