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Question
If a hexagon ABCDEF circumscribe a circle, prove that AB + CD + EF = BC + DE + FA.
Solution
According to the question,
A Hexagon ABCDEF circumscribe a circle.
To prove: AB + CD + EF = BC + DE + FA
Proof: Tangents drawn from an external point to a circle are equal.
Hence, we have
AM = RA ...Equation 1 [tangents from point A]
BM = BN ...Equation 2 [tangents from point B]
CO = NC ...Equation 3 [tangents from point C]
OD = DP ...Equation 4 [tangents from point D]
EQ = PE ...Equation 5 [tangents from point E]
QF = FR ...Equation 6 [tangents from point F] [equation 1] + [equation 2] + [equation 3] + [equation 4] + [equation 5] + [equation 6]
AM + BM + CO + OD + EQ + QF = RA + BN + NC + DP + PE + FR
On rearranging, we get,
(AM + BM) + (CO + OD) + (EQ + QF) = (BN + NC) + (DP + PE) + (FR + RA)
AB + CD + EF = BC + DE + FA
Hence Proved!
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