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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)?

#### question_answer 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) SME Approved

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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