Question

Show that in a quadrilateral ABCD, AB+BC+ CD+DA‹2(BD+AC)​

Answer 1

Step-by-step explanation:

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) 

Answer 2

Step-by-step explanation:

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