In the figure, ABCDEF is a regular hexagon. Prove that ΔBDF, drawn joining alternate vertices is equilateral.

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.