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

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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 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)

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