Question

In a quadrilateral ABCD in which diagonal AC & BD intersect at O. Show that AB + BC + CD + DA< 2(AC + BD)

Answer 1

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)

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Answer 2

Sol: In ABC , By inequality property of triangle,

AB + BC > AC -------(1)

In BCD , By inequality property of triangle,

BC + CD > BD -------(2)

In ADC , By inequality property of triangle,

CD + DA > AC -------(3)

In ABD , By inequality property of triangle,

DA + AB > BD -------(4)Add equation (1)(2)(3) and (4)

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

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

AB+BC+CD+DA > AC+BD

Hence Proved