Question
In a quadrilateral ABCD, show that
(AB + BC + CD + DA) <2 (BD + AC).
(AB + BC + CD + DA) <2 (BD + AC).
✓
Answer
Given: Quadrilateral ABCD
To prove: (AB + BC + CD + DA) < 2(BD + AC).
Proof:
In $\triangle\text{AOB},$
$\text{OA + OB > AB}...\text{(i)}$
In $\triangle\text{BOC},$
$\text{OB + OC > BC}...(\text{ii})$
In $\triangle\text{COD},$
$\text{OC + OD > CD}...(\text{iii})$
In $\triangle\text{AOD},$
$\text{OD + OA >AD}...(\text{iv})$
Adding (i), (ii), (iii) and (iv) we get
$2(\text{OA + OB + OC + OD})>(\text{AB + BC + CD + DA})$
$\Rightarrow2(\text{OB + OD + OA + OC})>(\text{AB + BC + CD + DA})$
$\Rightarrow2(\text{BD + AC})>(\text{AB + BC + CD + DA})$
To prove: (AB + BC + CD + DA) < 2(BD + AC).
Proof:
In $\triangle\text{AOB},$
$\text{OA + OB > AB}...\text{(i)}$
In $\triangle\text{BOC},$
$\text{OB + OC > BC}...(\text{ii})$
In $\triangle\text{COD},$
$\text{OC + OD > CD}...(\text{iii})$
In $\triangle\text{AOD},$
$\text{OD + OA >AD}...(\text{iv})$
Adding (i), (ii), (iii) and (iv) we get
$2(\text{OA + OB + OC + OD})>(\text{AB + BC + CD + DA})$
$\Rightarrow2(\text{OB + OD + OA + OC})>(\text{AB + BC + CD + DA})$
$\Rightarrow2(\text{BD + AC})>(\text{AB + BC + CD + DA})$
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