Question

if a hexagon ABCDEF circumscribe a circle. prove that
AB+CD+EF=BC+DE+AF

Answer 1

Sol:
Let ABCDEF be a hexagon.It is circumscribe a circle.

But length of the tangents drawn from a point to the circle are equal.

AM = AR -----(i)
BM = BN ---------(ii)
CN = CO ----------(iii)
DO = DP ----------(iv)
EP = EQ ----------(v)
FQ = FR --------- (vi)

adding (i) and (ii) we get 

AM + BM = AR + BN

⇒ AB = AR + BN

adding (iii) and (iv) we get 

CO + DO = CN + DP

⇒ CD = CN + DP

adding (v) and (vi) we get 

EQ + FQ = EP + FR

⇒ EF = EP + FR

Adding all these we obtain

AB + CD + EF = AR + ( BN + CN ) + (DP + EP) + FR = BC + DE + FA

∴ AB + CD + EF = BC + DE + FA