In the adjoining figure, ABCD is a quadrilateral and AC is one of its diagonals.

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

 

Solution:

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 ∆ABCAB + BC > AC            ...(1)

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

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