Question

Question 4
ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD?

Answer 1

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.
Therefore, In $\Delta$ ABC,AB + BC>AC ……….(i)
In $\Delta$ADC, AD + DC>AC ……….(ii)
In $\Delta$DCB, DC + CB>DB ……….(iii)
In $\Delta$ADB, AD + AB>DB ……….(iv)
Adding eq. (i), (ii), (iii) and(iv),
AB + BC + AD + DC + DC + CB + AD + AB > AC + AC + DB +DB
$\Rightarrow$ (AB + AB) + (BC + BC) + (AD + AD) + (DC + DC) > 2AC +2DB
$\Rightarrow$ 2AB + 2BC + 2AD + 2DC > 2(AC +DB)
$\Rightarrow$ 2(AB + BC + AD + DC) > 2(AC +DB)
$\Rightarrow$ AB + BC + AD + DC > AC +DB
$\Rightarrow$ AB + BC + CD + DA > AC +DB
Hence, it is true.