Question

Let ABCD be a quadrilateral in which the diagonals intersect at O perpendicularly. Prove that AB + BC + CD+ DA >AC + BD.

Answer 1

Given :

ABCD is a quadrilateral .

AC and BD are two diagonals .

To prove :

AB + BC + CD + DA > AC + BD

Proof :

i ) In ∆ABC ,

AB + BC > AC ---( 1 )

ii ) In ∆ADC ,

AD + DC > AC ---( 2 )

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The sum of any two sides of a

triangle is greater than the third

side .

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iii ) In ∆BCD ,

BC + CD > BD ---( 3 )

iv ) In ∆ADB ,

AB + AD > BD -----( 4 )

Add ( 1 ) ,( 2 ) , (3 ) and ( 4 ),we get

AB + BC + AD+DC+BC+CD+AB+AD >

AC + AC + BD + BD

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

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

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