Question

In the adjoining figure, ABCD is a quadrilateral and AC is one of its diagonals. Prove that:
(i) AB + BC + CD + DA > 2AC
(ii) AB + BC + CD > DA
(iii) AB + BC + CD + DA > AC + BD

Answer 1

Given: ABCD is a quadrilateral and AC is one of its diagonal.

(i) We know that the sum of any two sides of a triangle is greater than the third side.
In ∆ABC, AB + BC > AC ...(1)

In ∆ACD, CD + DA > AC ...(2)

​Adding inequalities (1) and (2), we get:

AB + BC + CD + DA > 2AC

(ii) In ∆ABC, we have :
​ AB + BC > AC ...(1)
We also know that the length of each side of a triangle is greater than the positive difference of the length of the other two sides.
In ∆ACD, we have:​
AC > |DA − CD|​ ...(2)
From (1) and (2), we have:
AB + BC > |DA − CD|​
⇒ AB + BC + CD > DA

(iii) In ∆ABC, AB + BC > AC

In ∆ACD, CD + DA > AC

In ∆ BCD, BC + CD > BD

In ∆ ABD, DA + AB > BD
​Adding these inequalities, we get:
2(AB + BC + CD + DA) > 2(AC + BD)
⇒ (AB + BC + CD + DA) > (AC + BD)