# If a hexagon ABCDEF circumscribe a circle, prove thatAB + CD + EF = BC + DE + FA 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  [tangents from point A]

BM = BN  [tangents from point B]

CO = NC  [tangents from point C]

OD = DP  [tangents from point D]

EQ = PE  [tangents from point E]

QF = FR  [tangents from point F]

Add      and 

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 !

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