Q. 14.2( 20 Votes )

If a hexagon ABCDEF circumscribe a circle, prove that

AB + CD + EF = BC + DE + FA

Answer :


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 !


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