Question

In a quadrilateral $ABCD$ show that $AB+BC+CD+DA<2(BD+AC)$

Answer 1

Construction: Join diagonals $AC$ and $BD$



In $\Delta OAB,OA+OB>AB$ ... (i)
[sum of any two sides of a triangle is greater than the third side]
In $\Delta OBC,OB+OC>BC$ ... (ii)
[sum of any two sides of a triangle is greater than the third side]
Similarly, in $\Delta OCD$,
$OC+OD>CD$ ... (iii)
And, in $\Delta ODA$,
$OD+OA>DA$ ... (iv)
On adding eqs (i), (ii), (iii) and (iv), we get
$2\left[\right(OA+OB+OC+OD]>AB+BC+CD+DA$
$\Rightarrow 2\left[\right(OA+OC)+(OB+OD\left)\right]>AB+BC+CD+DA$
$\Rightarrow 2(AC+BD)>AB+BC+CD+DA$
$[\because OA+OC=AC$ and $OB+OD=BD]$
$\Rightarrow AB+BC+CD+DA<2(BD+AC)$