Question

Let ABCD be a quadrilateral with diagonals AC and BD. Prove the following statements
(a) AB + BC + CD > AD
(b) AB + BC + CD + DA > 2AC(c) AB + BC + CD + DA > 2BD
(d) AB + BC + CD + DA > AC + BD.

Answer 1

Solution :

i )
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sum of any two sides of a

triangle is greater than the third side

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so

In ∆ABC , AB + BC > AC ----- ( 1 )

In ∆ACD , AC + CD > DA ---- ( 2 )

Adding ( 1 ) and ( 2 ) , we get

AB + BC + AC + CD > AC + DA

=> AB + BC + AC + CD > AC + DA

=> AB + BC + CD > DA

ii ) In ∆ACD , CD + DA > AC ---- ( 3 )

Adding ( 1 ) and ( 3 ) , we get

AB + BC + CD + DA > 2AC ----( 4 )

iii ) In ∆ABD, DA + AB > BD -----( 5 )

In ∆BCD , BC + CD > BD -------( 6 )

Adding ( 5 ) and ( 6 ), we get

AB + BC + CD + DA > 2BD ---( 7 )

iv ) Adding ( 4 ) and ( 7 ) , we get

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

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

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