In the figure below, ABCDEF is a regular hexagon. Prove that ∆ BDF i

1.	In the figure below, ABCDEF is a regular hexagon. Prove that ∆ BDF is an equilateral triangle?
Answer» Sum of angles of a regular hexagon = (6 – 2) 180 = 720° = 4 × 180 = 720° One angle = \(\frac{720}{6}\) = 120° consider ∆ EFO ∠E = 120° ∠EFD = ∠EDF = 30°(Angles opposite to equal sides of an isosceles triangle are equal) Similarly ∠AFB = 30° ∴∠DFB = 120 – (30 + 30) = 60° ∴ ∠FBD = 60°, ∠FDB = 60° ∴ ∆ FDB is an equilateral triangle.

In the figure below, ABCDEF is a regular hexagon. Prove that ∆ BDF is an equilateral triangle?

Answer»

Sum of angles of a regular hexagon = (6 – 2) 180 = 720°

= 4 × 180 = 720°

One angle = \(\frac{720}{6}\) = 120°

consider ∆ EFO

∠E = 120°

∠EFD = ∠EDF = 30°(Angles opposite to equal sides of an isosceles triangle are equal)

Similarly ∠AFB = 30°

∴∠DFB = 120 – (30 + 30) = 60°

∴ ∠FBD = 60°,

∠FDB = 60°

∴ ∆ FDB is an equilateral triangle.