Question

please prove the attached question​

Answer 1

your answer is here.

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

sum of the two sides of a triangle is greater than the third side.

therefore,

in triangle.. ABC,

AB+BC > AC ..........(i)

in triangle.. BCD,

BC+CD > BD ..........(ii)

in triangle.. CDA,

CD+DA > AC ..........(iii)

in triangle.. DAB,

DA+AB > BD ..........(iv)

adding (i), (ii), (iii), (iv), we get ,,,

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

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

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

::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

ii). (AB+BC+CD+DA) < 2(AC+BD)

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 proved.

i hope it helps you..

make as brilliant...

thank you..!!