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