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ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD ?

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

Therefore,

In Δ ABC, AB + BC > AC ……….(i) 

In Δ ADC, AD + DC > AC ……….(ii) 

In Δ DCB, DC + CB > DB ……….(iii) 

In Δ ADB, AD + AB > DB ……….(iv)

Adding eq. (i), (ii), (iii) and (iv),

 AB + BC + AD + DC + DC + CB + AD + AB > AC + AC + DB + DB 

⇒ (AB + AB) + (BC + BC) + (AD + AD) + (DC + DC) > 2AC + 2DB 

⇒ 2AB + 2BC + 2AD + 2DC > 2(AC + DB) 

⇒ 2(AB + BC + AD + DC) > 2(AC + DB) 

⇒ AB + BC + AD + DC > AC + DB 

⇒ AB + BC + CD + DA > AC + DB 

Hence, it is true.

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