Math, asked by vaidangvyas, 1 year ago

ABCD is a quadrilateral in which diagonal AC and BD intersect at O. show that AB+BC+CD+DA > AC + BD

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Answered by soumyakapis35pdtdz0
5
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Answered by Anonymous
8

Answer:

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

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