Question

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

Answer 1

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.