In the figure, ABCDEF is a regular hexagon. Prove that ACDF is a rectangle.

In Δ ABC and Δ FED

AB = FE [sides of hexagon]

BC = ED [sides of hexagon]

∠ ABC = ∠ FED = 120°

∴ Δ ABC ≅ Δ FED

AC = FD [CPCT]

In Δ ABC

∠ BAC = ∠ BCA [angle made on opposite sides]

∠ BAC + ∠ BCA + ∠ ABC = 180°

2∠ BCA + 120° = 180°

⇒ 2 ∠ BCA = 180° - 120°

⇒ 2∠ BCA = 60°

⇒

∴ ∠ BCA = ∠ BAC = 30°

Similarly, ∠ EFD = ∠ EDF = 30°

∠ BCD = ∠ BCA + ∠ ACD

⇒ 120° = 30° + ∠ ACD

⇒ ∠ ACD = 120° - 30°

⇒ ∠ ACD = 90°

∠ EDC = ∠ FDE + ∠ FDC

⇒ 120° = 30° + ∠ FDC

⇒ ∠ FDC = 120° - 30°

⇒ ∠ FDC = 90°

∴ From above calculations AC = FD and AF = CD and ∠ FDC = ∠ ACD = 90°

∴ ACDF is a rectangle.