Question

In a quad. show that (AB+BC+CD+AD) < 2(AC+BD) ​

Answer 1

Answer:

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

Too prove opposite side of an quadratel are supplementry

Proof. Angle 12345678 are equal

Angle 8=1

Similarly

Angle. 2=3

4=5

6=7

Hence proved