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 !