Question

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

Answer 1

Answer:

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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

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

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third sideTherefore, In Δ AOB, AB < OA + OB ……….(i)

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third sideTherefore, In Δ AOB, AB < OA + OB ……….(i) In Δ BOC, BC < OB + OC ……….(ii)

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third sideTherefore, In Δ AOB, AB < OA + OB ……….(i) In Δ BOC, BC < OB + OC ……….(ii) In Δ COD, CD < OC + OD ……….(iii)

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

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

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third sideTherefore, 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)]

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

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

Answer:

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