If a hexagon ABCDEF circumscribe a circle, prove that AB + CD + EF = BC + DE + FA.
Solution
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!
