Question

Question 6
(a) In a quadrilateral ABCD, prove that AB + BC + CD + DA > 2BD. Draw the Quad also if possible.​

Answer 1

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.