Given : A Hexagon ABCDEF circumscribe a circle .
To prove : AB + CD + EF = BC + DE + FA
Proof :
As we know, that
Tangents drawn from an external point to a circle are equal.
We have
AM = RA [1] [tangents from point A]
BM = BN [2] [tangents from point B]
CO = NC [3] [tangents from point C]
OD = DP [4] [tangents from point D]
EQ = PE [5] [tangents from point E]
QF = FR [6] [tangents from point F]
Add [1] [2] [3] [4] [5] and [6]
AM + BM + CO + OD + EQ + QF = RA + BN + NC + DP + PE + FR
Rearranging
(AM + BM) + (CO + OD) + (EQ + QF) = (BN + NC) + (DP + PE) + (FR + RA)
AB + CD + EF = BC + DE + FA
Hence Proved !
