Q. 84.0( 6 Votes )

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Answer :

Construction: Draw a perpendicular AO on BF


All the angles are 120° as it is regular hexagon and all the sides are equal.


Now in Δ AOB and Δ AOF,


AO = AO [common]


AB = AF [all sides are equal]


AOB = AOF = 90° [by construction]


So, Δ AOB Δ AOF


BAO = FAO = 60° [CPCT]


ABO = AFO [CPCT] … (1)


Similarly, Δ BGC Δ DGC


BCG = DCG = 60°


CBG = CDG … (2)


Similarly, Δ DHE Δ FHE


DEH = FEH = 60°


EDH = EFH … (3)


In Δ BAO,


BAO + AOB + ABO = 180°


60° + 90° + ABO = 180°


150° + ABO = 180°


ABO = 180° - 150°


ABO = 30°


from (1) ABO = AFO = 30°


Similarly


CBG = CDG = 30°


EDH = EFH = 30°


ABO + FBD + CBG = ABC


30° + FBD + 30° = 120°


FBD + 60° = 120°


FBD = 120° - 60°


FBD = 60°


Similarly,


BDF = DFB = 60°


In Δ BDF


FBD = BDF = DFB = 60°


Δ BDF is an equilateral triangle.


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