To divide a regular hexagon into six congruent equilateral triangles, start by understanding the geometry of the hexagon and its relationship with the circle in which it is inscribed.
A. Drawing Line Segments
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Draw Diagonals from One Vertex:
- Choose one vertex (let's call it \( A \)) of the hexagon.
- Draw line segments from vertex \( A \) to the other non-adjacent vertices (let's denote them as \( C \) and \( E \)). This will create two triangles: \( \triangle ABC \) and \( \triangle AEF \). The lines \( AC \) and \( AE \) divide the hexagon into three triangles in total.
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Connect Adjacent Vertices:
- Next, draw line segments connecting the vertices of the hexagon in a way that creates more triangles. Specifically, connect vertex \( B \) to vertex \( D \) and vertex \( C \) to vertex \( F \). This will create additional line segments that can help define triangles.
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Use the Center:
- Draw lines from the center of the hexagon (which is also the center of the circumcircle) to each of the vertices \( A, B, C, D, E, F \). These segments will intersect with the sides of the hexagon, dividing it into six triangles (\( \triangle OAB, \triangle OBC, \triangle OCD, \triangle ODE, \triangle OEF, \triangle OFA \)).
B. Explanation of Equilateral Triangles
To prove that each triangle created in part A is equilateral, we can use the properties of the regular hexagon and the fact that it is inscribed in a circle.
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Equal Side Lengths:
- In a regular hexagon, all sides are of equal length. Hence, any triangle formed by connecting the vertices of the hexagon will have at least two sides equal to the length of a side of the hexagon. For example, in triangle \( OAB \), the sides \( OA \) and \( OB \) are equal (both are radii of the circumcircle).
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Equal Angles:
- The central angle subtended by each side of the hexagon (from the center \( O \) to adjacent vertices, like \( A \) and \( B \)) is \( 60^\circ \) since there are six equal angles in a complete circle. Hence, the angles \( \angle AOB, \angle BOC, \angle COD, \angle DOE, \angle EOF, \angle FOA \) are all \( 60^\circ \).
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Using the Radius:
- Each triangle (like \( OAB \)) has sides \( OA, OB, \) and \( AB \), where \( OA = OB \) (radii), and \( AB = \text{side of the hexagon} \). Since each angle of these triangles is \( 60^\circ \), the triangles formed are equilateral.
Thus, you can conclude that each of the six triangles formed by the divisions is indeed equilateral since they are bounded by equal sides of the hexagon and angles of \( 60^\circ \).