136
If a regular hexagon ABCDEF circumscribes a circle, then prove that $AB + CD + EF = BC + DE + FA$.
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Sol. correct figure (1/2 Mark)
AM = AR
BM=BN
CN = CO
DO = DP
EQ=EP
FQ=FR$ (1 Mark)\LHS $= AB+CD+EF = (AM+BM) + (CO+OD) + (EQ + QF)$ (1/2 Mark)\\$= AR+ BN + CN + DP + EP + FR$ (1/2 Mark)\\$= (AR+FR) + (BN+CN)+ (DP+EP)$ (1/2 Mark)\\$= BC + DE+ FA = RHS$
AM = AR
BM=BN
CN = CO
DO = DP
EQ=EP
FQ=FR$ (1 Mark)\LHS $= AB+CD+EF = (AM+BM) + (CO+OD) + (EQ + QF)$ (1/2 Mark)\\$= AR+ BN + CN + DP + EP + FR$ (1/2 Mark)\\$= (AR+FR) + (BN+CN)+ (DP+EP)$ (1/2 Mark)\\$= BC + DE+ FA = RHS$