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Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD

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Question

Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD

Sum
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Solution

Given in the question, a quadrilateral ABCD.

To proof that AB + BC + CD + DA > AC + BD.

Proof: In triangle ABC,


AB + BC > AC   ...(i) [Sum of the lengths of any two sides of a triangle must be greater than the third side]

In triangle BCD,

BC + CD > BD   ...(ii) [Sum of the lengths of any two sides of a triangle must be greater than the third side]

In triangle CDA,

CD + DA > AC   ...(iii) [Sum of the lengths of any two sides of a triangle must be greater than the third side]

Similarly, in triangle DAB,

AD + AB > BD   ...(iv) [Sum of the lengths of any two sides of a triangle must be greater than the third side]

Now, adding equation (i), (ii), (iii) and (iv), we get

AB + BC + BC + CD + CD + DA + AD + AB > AC + BD + AC + BD

2AB + 2BC + 2CD > 2AC + 2BD

2(AB + BC + CD + DA) > 2(AC + BD)

AB + BC + CD + DA > AC + BD

Hence proved.

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Chapter 7: Triangles - Exercise 7.4 [Page 70]

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NCERT Exemplar Mathematics Exemplar [English] Class 9
Chapter 7 Triangles
Exercise 7.4 | Q 12. | Page 70

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