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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)?

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)

â‡’ 2(AB + BC + CA + AD) > 2(AC + BD)

â‡’ (AB + BC + CA + AD) > (AC + BD)

â‡’ (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

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)

â‡’ 2(AB + BC + CA + AD) > 2(AC + BD)

â‡’ (AB + BC + CA + AD) > (AC + BD)

â‡’ (AC + BD) < (AB + BC + CA + AD)

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)

â‡’ 2(AB + BC + CA + AD) > 2(AC + BD)

â‡’ (AB + BC + CA + AD) > (AC + BD)

â‡’ (AC + BD) < (AB + BC + CA + AD)

Amna javeed

Member since

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