Show that in a quadrilateral $ABCD$,$AB + BC + CD + DA < 2(BD + AC)$.

(N/A) Given: $A$ quadrilateral $ABCD$ with diagonals $AC$ and $BD$ intersecting at point $O$.
To prove: $AB + BC + CD + DA < 2(BD + AC)$.
Proof: In $\triangle AOB$,by the triangle inequality theorem (the sum of any two sides of a triangle is greater than the third side),we have:
$OA + OB > AB$ ... $(1)$
In $\triangle BOC$,we have:
$OB + OC > BC$ ... $(2)$
In $\triangle COD$,we have:
$OC + OD > CD$ ... $(3)$
In $\triangle DOA$,we have:
$OD + OA > DA$ ... $(4)$
Adding equations $(1), (2), (3),$ and $(4)$,we get:
$(OA + OB) + (OB + OC) + (OC + OD) + (OD + OA) > AB + BC + CD + DA$
$2(OA + OC) + 2(OB + OD) > AB + BC + CD + DA$
Since $OA + OC = AC$ and $OB + OD = BD$,we substitute these into the inequality:
$2(AC) + 2(BD) > AB + BC + CD + DA$
$2(AC + BD) > AB + BC + CD + DA$
Therefore,$AB + BC + CD + DA < 2(AC + BD)$.
Hence,proved.