Question

ABCD is a quadrilateral is AB + BC + CD + DA greater than 2 AC + BD​

Answer 1

$\huge\sf\blue{Answer:-}$

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.

$\boxed{Therefore,}$

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 proved.

Answer 2

heey dear

follow me

thank me