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Question
In a quadrilateral ABCD; prove that:
(i) AB+ BC + CD > DA
(ii) AB + BC + CD + DA > 2AC
(iii) AB + BC + CD + DA > 2BD
Solution
Const: Join AC and BD.
(i) In ΔABC,
AB + BC > AC ….(i)[ Sum of two sides is greater than the third side ]
In ΔACD,
AC + CD > DA ….(ii)[ Sum of two sides is greater than the third side ]
Adding (i) and (ii)
AB + BC + AC + CD > AC + DA
AB + BC + CD > AC + DA - AC
AB + BC + CD > DA ….(iii)
(ii) In ΔACD,
CD + DA > AC ….(iv) [Sum of two sides is greater than the third side]
Adding (i) and (iv)
AB + BC + CD + DA > AC + AC
AB + BC + CD + DA > 2AC
(iii) In ΔABD,
AB + DA > BD ….(v)[Sum of two sides is greater than the third side]
In ΔBCD,
BC + CD > BD ….(vi)[Sum of two sides is greater than the third side]
Adding (v) and (vi)
AB + DA + BC + CD > BD + BD
AB + DA + BC + CD > 2BD
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