Since, the sum of lengths of any two sides in a triangle should be greater than the length
of third side.
Therefore, In ∆ABC, AB + BC > AC ……….(i)
In ∆ADC, AD + DC > AC ……….(ii)
In ∆DCB, DC + CB > DB ……….(iii)
In ∆ADB, AD + AB > DB ……….(iv)
Adding equations (i), (ii), (iii) and (iv), we get
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.

Class 7 Maths Chapter 6 Exercise 6.4

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