Question

Abcd is a quadrilateral then prove that ab + bc + cd + da < 2(ac + bd)?

Answer 1

Answer:

2(ac + bd) is greater

please mark it as brainliest

Answer 2

Given:

ABCD is quadrilateral$.$

To find:

AB+BC+CD+DA<2(AC+BD) ?

STEP BY STEP EXPLANATION:

In a triangle, the sum of the lengths of either two sides is always greater than the third side.

Considering ∆ ABC,

AB + BC > CA (i)

In ∆ ABCD,

BC + CD > DB (ii)

In ∆ ACDA,

CD + DA > AC (iii)

In ∆ ADAB,

DA + AB > DB (iv)

Adding equations (i), (ii), (ii), and (iv), we obtain

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

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

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

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

∴ Yes, the given expression is true.

Hence verified !