A regular hexagon can be divided into six congruent equilateral triangles.

Question

On the paper where you constructed your regular hexagon inscribed in a circle, use your straightedge to draw three different line segments that divide your constructed hexagon into six equilateral triangles.

Explain how you know that each of these triangles is equilateral.

Answer 1

To show how to divide a regular hexagon into six congruent equilateral triangles and explain why each triangle is equilateral, we can follow these steps:

Step 1: Construct the Regular Hexagon

1. Draw a Circle: Start by drawing a circle of radius \( r \).
2. Mark the Vertices: Divide the circle into six equal parts, marking the vertices \( A, B, C, D, E, F \) clockwise. Each central angle between adjacent vertices is \( 60^\circ \).
3. Connect the Vertices: Connect adjacent vertices \( A \) to \( B \), \( B \) to \( C \), and so on, until \( F \) connects back to \( A \). This forms a regular hexagon.

Step 2: Draw the Segments

To divide the hexagon into six equilateral triangles, you can draw three specific line segments from the center of the hexagon, denoted as point \( O \) (the center of the circle).

1. Draw a Line from \( O \) to \( A \): This creates triangle \( OAB \).
2. Draw a Line from \( O \) to \( C \): This creates triangle \( OCD \).
3. Draw a Line from \( O \) to \( E \): This creates triangle \( OEF \).

The same procedure is followed for the rest of the triangles \( OBC, OCD, OEF, \) and \( OFA \).

Step 3: Show All Triangles Are Equilateral

To prove that each triangle \( OAB, OBC, OCD, ODE, OEF, OFA \) is equilateral, we need to demonstrate that all its sides are equal:

1. All Sides are Radii of the Circle:

   • Since the hexagon is inscribed in a circle, each vertex \( A, B, C, D, E, F \) lies on the circle.
   • The segments \( OA, OB, OC, OD, OE, OF \) are radii of the circle, which are all equal to the radius \( r \) of the circle.

2. Angles:

   • The central angles subtended by each side of the hexagon at point \( O \) is \( 60^\circ \) (since \( 360^\circ / 6 = 60^\circ \)).
   • Each triangle \( OAB, OBC, OCD, ODE, OEF, OFA \) thus has one angle of \( 60^\circ \) at the vertex \( O \).

3. Using the Triangle Properties:

   • In triangle \( OAB \):
     • \( OA = OB = r \) (both are equal radii).
     • The angle \(\angle AOB = 60^\circ\).
     • By the properties of triangles, with two sides equal and the included angle being \( 60^\circ \), triangle \( OAB \) is equilateral.

The same reasoning applies to all other triangles formed (i.e., \( OBC, OCD, ODE, OEF, OFA \)), establishing that they are also equilateral.

Conclusion

Thus, by drawing segments from the center \( O \) to the vertices of the hexagon, you divide the hexagon into six congruent equilateral triangles, confirming that each triangle is equilateral due to having all side lengths equal (as they are all radii of the circle) and each angle being \( 60^\circ \).