Prove if a circle is divided into n congruent arcs (n \ ge 3), the chords determined by joining consecutive end points of these arcs they form a regular polygon?

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If the radius of the circle is r, its circumference is 2πr, so the length of each arc is 2πr/n. When the endpoints of the arc are joined to the centre of the circle they form radii. Each arc subtends an angle 2π/n at the centre of the circle. The chord joining the endpoints and the two radii form an isosceles triangle with vertex angle 2π/n radians (360/n degrees). The equal angles measure ½(π-2π/n) radians=½π-π/n radians (90-180/n degrees). Since the arcs are congruent, so are the isosceles triangles all round the circle. The angle between two consecutive chords is twice the measure of the equal angles of the isosceles triangles=π-2π/n radians (180-360/n degrees).

The interior angles of an n-sided polygon (regular or irregular) sum to 2n-4 right angles, that is, (2n-4)π/2=nπ-2π radians or 180n-360 degrees. Each interior angle of a regular polygon (all interior angles equal in measure) therefore measures (nπ-2π)/n=π-2π/n radians or (180n-360)/n=180-360/n degrees. But this is the same measure as the angle between two consecutive chords formed by the congruent arcs. Therefore, the chords form the sides of the regular polygon.

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