Question

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

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 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

Joining A to C and B to D, we get traiangles, ABC, BDC, ACD and ABD.  We know that sum of any two sides of a triangle is greater than the third side.  So, let us apply the above property stated and find the required result  In triangle ABC;  AB + BC > AC In triangle ABD;  AB + AD > BD In triangle DBC;  CD + BC > BD and In triangle ADC;  AD + DC > AC Adding all the above inequalities, we get 2(AB + BC + AD + DC) > 2(AC + BD) i.e, AB + BC + AD + DC > AC + BD