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Okay, so the sum says

In a quadrilateral ABCD prove that
can anyone help with this...??? Thanks:)



Now by the property of triangles, sum of any two sides > the third side. 
Therefore, AB + BC > AC Same way CD +DA > AC 
Similar way we can get DA + AB > BD and BC + CD > BD 
Now adding all these inequalities, 2(AB + BC +CD +DA) > 2(AC + BD) 
Dividing both sides by 2 we get, (AB + BC +CD +DA) > (AC + BD) 
Thus, the given first identity is proved. 
For the second identity, 
2AC > AB + BC Similar way, 2AC > CD + DA 
Also, 2BD > DA + AB and 2BD > BC + CD 
Adding the inequalities, we get 4(AC + BD) > 2(AB + BC +CD +DA) 
Dividing by 2 on both sides, 
we arrive at 2(AC + BD) > (AB + BC +CD +DA) 
or (AB + BC +CD +DA) < 2(AC + BD) Hence second identity is proved.
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