Question

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

Answer 1

Given: Quadrilateral ABCD
To prove: (AB + BC + CD + DA ) < 2( BD + AC ).
Proof:
In ∆AOB,
OA+OB>AB------i
In ∆BOC,
OB+OC>BC-----ii
In ∆COD,
OC+OD>CD----iii
In ∆AOD,
OD+OA>AD-----iv
Adding i,ii,iii and iv, we get
2OA+OB+OC+OD>AB+BC+CD+DA⇒ 2OB+OD+OA+OC>AB+BC+CD+DA⇒2BD+AC>AB+BC+CD+DA