In a quadrilateral ABCD, show that (AB + BC + CD + DA) <2 (BD + A… - Vidyadip

Question

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

✓

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 $2(\text{OA + OB + OC + OD})>(\text{AB + BC + CD + DA})$
$\Rightarrow2(\text{OB + OD + OA + OC})>(\text{AB + BC + CD + DA})$
$\Rightarrow2(\text{BD + AC})>(\text{AB + BC + CD + DA})$

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