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Asked by Principal Mahadevi beede

Dec 29, 2014

ABCD is a quadrilateral then prove that AB + BC + CD + DA < 2(AC + BD)?

ABCD is a quadrilateral then prove that AB + BC + CD + DA < 2(AC + BD)?

Answers(4)

Answer

Syeda

Member since Jan 25, 2017

Answer.

ABCD is a quadrilateral and AC, and BD are the diagonals.
Sum of the two sides of a triangle is greater than the third side.
So, considering the triangle ABC, BCD, CAD and BAD, we get
AB + BC > AC
CD + AD > AC
AB + AD > BD
BC + CD > BD

Adding all the above equations,

2(AB + BC + CA + AD) > 2(AC + BD)
⇒ 2(AB + BC + CA + AD) > 2(AC + BD)
⇒ (AB + BC + CA + AD) > (AC + BD)
⇒ (AC + BD) < (AB + BC + CA + AD)

Ashok K Singh

Member since

Joining A to C and B to D, we get traiangles, ABC, BDC, ACD and ABD. 
We know that sum of any two sides of a triangle is greater than the third side. 
So, let us apply the above property stated and find the required result 
In triangle ABC;  AB + BC > AC
In triangle ABD;  AB + AD > BD
In triangle DBC;  CD + BC > BD
and In triangle ADC;  AD + DC > AC
Adding all the above inequalities, we get

2(AB + BC + AD + DC) > 2(AC + BD)
i.e, AB + BC + AD + DC > AC + BD
 

Ramesh

Member since Apr 1, 2014

ABCD is a quadrilateral and AC, and BD are the diagonals.
Sum of the two sides of a triangle is greater than the third side.
So, considering the triangle ABC, BCD, CAD and BAD, we get
AB + BC > AC
CD + AD > AC
AB + AD > BD
BC + CD > BD

Adding all the above equations,

2(AB + BC + CA + AD) > 2(AC + BD)
⇒ 2(AB + BC + CA + AD) > 2(AC + BD)
⇒ (AB + BC + CA + AD) > (AC + BD)
⇒ (AC + BD) < (AB + BC + CA + AD)

Amna javeed

Member since

Not necessary. The answer will be according to the given sides.
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