Question

ABCD is quadrilateral.
Is AB+BC+CD+DA<2(AC+BD) ?

Answer 1

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side Therefore,  In Δ AOB, AB < OA + OB ……….(i)  In Δ BOC, BC < OB + OC ……….(ii)  In Δ COD, CD < OC + OD ……….(iii)  In Δ AOD, DA < OD + OA ……….(iv)  ⇒ AB + BC + CD + DA < 2OA + 2OB + 2OC + 2OD  ⇒ AB + BC + CD + DA < 2[(AO + OC) + (DO + OB)]  ⇒ AB + BC + CD + DA < 2(AC + BD)  Hence, it is prove.

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 AABC,

AB + BC > CA ()

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 +

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 !