Draw the perpendicular bisectors of the line segments

AB

and

BC

, then the intersection of the perpendicular bisectors is equidistant from

A

B

and

C

and thus it is the centre of the circle with

AB

and

BC

as chords. All the perpendicular bisectors will meet in a single point

O

equidistant from the endpoints of the line segments. Therefore all the line segments are chords of a single circle with centre

O

The angle of turn between the equal chords

AB

and

BC

in the path is

∠ PBC = θ

. Triangles

AOB

and

BOC

are isosceles and

∠ OAB = ∠ OBA = ∠ OBC = α

. Then (using angles on a straight line and angles in a triangle)

θ = 180 - 2 α = ∠ AOB

Consider radii drawn from the centre to the ends of

n

consecutive chords in the path. The sum of angles between the radii is equal to the total angle turn

n θ

so the path is closed if and only if the total angle turn is an integer multiple of 360 degrees, that is if and only if

n θ = 360 m

or, equivalently

θ

is a rational multiple of 360 degrees.

We consider $θ = 360 \frac{p}{q}$ where $p$ and $q$ are positive integers and $p < q$ . If the angle of turn is more than 360 degrees, complete revolutions can be discounted, so effectively $θ = 360 \frac{p *}{q}$ produces the same path as $θ = 360 \frac{p}{q}$ where $p * \equiv p$ mod $q$ .

If $p$ is a factor of q, (say $q = kp$ ) then the path is a regular $k$ sided polygon because the total angle turned when $k$ chords are drawn is $k \times \frac{360}{k} = 360$ degrees.

If $p$ and $q$ are not coprime then $sp = tq$ for some integers $s$ and $t$ so that the total angle turn with s chords is $s \times 360 \frac{p}{q} = 360 t$ which is an integer multiple of 360 degrees so the path is a star with $s$ points. If $p$ and $q$ are coprime then the path is a $q$ pointed star.