Question

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

Answer 1

Given: Quadrilateral ABCD
To prove: (AB + BC + CD + DA) < 2(BD + AC).
Proof:

In ∆AOB,
$OA+OB>AB.....\left(i\right)$
In ∆BOC,
$OB+OC>BC.....\left(ii\right)$
In ∆COD,
$OC+OD>CD.....\left(iii\right)$
In ∆AOD,
$OD+OA>AD.....\left(iv\right)$
Adding $\left(i\right),\left(ii\right),\left(iii\right)\mathrm{and}\left(iv\right)$, we get
$$\begin{gathered} 2\left(OA+OB+OC+OD\right)>\left(AB+BC+CD+DA\right) \\ \Rightarrow 2\left(OB+OD+OA+OC\right)>\left(AB+BC+CD+DA\right) \\ \Rightarrow 2\left(BD+AC\right)>\left(AB+BC+CD+DA\right) \end{gathered}$$