Question
In a quadrilateral ABCD, show that (AB + BC + CD + DA) > (AC + BD).
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Answer
Given: Quadrilateral ABCD To prove: (AB + BC + CD + DA) > (AC + BD) Proof: 
In $\triangle\text{ABC},$$\text{AB + BC > AC }...(\text{i})$
In $\triangle\text{CAD},$$\text{CD + AD > AC }...(\text{ii})$
In $\triangle\text{BAD,}$$\text{AB + AD + BD } ...(\text{iii})$
In $\triangle\text{BCD},$$\text{BC + CD + BD }...(\text{iv})$
Adding (i), (ii), (iii) and (iv), we get$2(\text{AB + BC + CD + DA})< 2 \text{(AC + BD)}$
Hence, $\text{(AB + BC + CD + DA)}<(\text{AC + BD}).$
In $\triangle\text{ABC},$$\text{AB + BC > AC }...(\text{i})$In $\triangle\text{CAD},$$\text{CD + AD > AC }...(\text{ii})$
In $\triangle\text{BAD,}$$\text{AB + AD + BD } ...(\text{iii})$
In $\triangle\text{BCD},$$\text{BC + CD + BD }...(\text{iv})$
Adding (i), (ii), (iii) and (iv), we get$2(\text{AB + BC + CD + DA})< 2 \text{(AC + BD)}$
Hence, $\text{(AB + BC + CD + DA)}<(\text{AC + BD}).$
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