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