Question
Show that in a quadrilateral $ABCD, AB + BC + CD + DA < 2 (BD + AC)$
✓
Answer
Given, A quadrilateral $ABCD.$
To prove: $AB + BC + CD + DA < 2(BD + AC)$
Proof: In $\triangle\text{AOD}$ We have $\therefore\text{OA}+\text{OB}>\text{AB }...(\text{i}) $
$\big[\therefore$ Sum of the lengths of any sides of a triangle must be greater than rhird side$\big]$
In $\triangle\text{BOC}$ We have $OB + OC + > BC ...(2) [$Same reason$]$
In $\triangle\text{COD},$ We have $OC + OD + > CD ...(3) [$Same reason$]$
In $\triangle\text{DOA},$ We have $OD + OA > DA ...(4) [$Same reason$]$
Adding $(1), (2), (3)$ and $(4),$
We get $OA + B + OB + OC + OC + OD + OA + > AB + BC + CD +DA $
$\Rightarrow 2(OA + OB + OC + OD) > AB + BC + CD + DA $
$\Rightarrow 2{(OA + OC) + (OB + OD)} > AB + BC + CD + DA $
$\Rightarrow 2(AC + BD) > AB + BC + CD + DA $
$\Rightarrow AB + BC + CD + DA < 2(BD + AC)$
Hence, proved.
To prove: $AB + BC + CD + DA < 2(BD + AC)$
Proof: In $\triangle\text{AOD}$ We have $\therefore\text{OA}+\text{OB}>\text{AB }...(\text{i}) $
$\big[\therefore$ Sum of the lengths of any sides of a triangle must be greater than rhird side$\big]$
In $\triangle\text{BOC}$ We have $OB + OC + > BC ...(2) [$Same reason$]$
In $\triangle\text{COD},$ We have $OC + OD + > CD ...(3) [$Same reason$]$
In $\triangle\text{DOA},$ We have $OD + OA > DA ...(4) [$Same reason$]$
Adding $(1), (2), (3)$ and $(4),$
We get $OA + B + OB + OC + OC + OD + OA + > AB + BC + CD +DA $
$\Rightarrow 2(OA + OB + OC + OD) > AB + BC + CD + DA $
$\Rightarrow 2{(OA + OC) + (OB + OD)} > AB + BC + CD + DA $
$\Rightarrow 2(AC + BD) > AB + BC + CD + DA $
$\Rightarrow AB + BC + CD + DA < 2(BD + AC)$
Hence, proved.
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