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.