Question

In the figure below, ABCDEF is regular hexagon.

Prove that ACE is an equilateral triangle.

Answer 1

The given figure is a regular hexagon, so all its angles are equal in measure.

Number of sides of a regular hexagon = 6

We know that the sum of the angles of a polygon with n sides = (n −2) × 180°

Sum of the angles of the regular hexagon = (6 − 2) × 180°

= 4 × 180°

= 720°

Let the measure of each angle of the given regular hexagon be x°.

⇒ 6x = 720°

⇒ x = 120°

In ΔFEA, by angle sum property, we have:

∠AFE +∠FEA + ∠EAF = 180° …(1)

FE = FA (All sides of a regular hexagon are equal in length)

⇒∠FEA = ∠FAE (Angles opposite to equal sides are equal in measure)

Putting in equation (1):

∠AFE + 2∠FEA = 180°

⇒ 120° + 2∠FEA = 180°

⇒ 2∠FEA = 60°

⇒∠FEA = 30°

Similarly, ∠DEC = ∠DCE = ∠BCA = ∠BAC = ∠FAE = 30°

As ∠FED = 120°

⇒ ∠FEA + ∠AEC + ∠CED = 120°

⇒ 30° + ∠AEC + 30° = 120°

⇒ 60° + ∠AEC = 120°

⇒∠AEC = 60°

Similarly, ∠ACE = ∠EAC = 60°

Thus, ΔAEC is an equilateral triangle.