You can construct a regular dodecagon or twelve sided figure by cutting off the corners of a regular hexagon. The question is how much do you cut off of each corner. If we let the sides of the regular hexagon each be one unit and we let the amount we cut off the corners be x, then we get the following figure.
To find x, we note that x satisfies the following equation.
To solve for x we transpose the x terms to the same side of the equation,
combine like terms,
and divide both sides by the coefficient of x.
Rationalize the denominator.
It is easy to find a 2 and a square root of 3 in the regular hexagon.
Now that we have the distance we need to take off of the corners, we can mark it off from each vertex.
To get a geometric proof that this construction works, we would need to show that in a regular dodecagon, HIJKLMNPQRST, where G is the point on the diameter, AD, whose distance from A is the same as the length of the side of the regular dodecagon, that B and G are the same distance from D. This is the same thing as saying that triangle GBD is isosceles. Since /BDG is a 30o angle, in order for the triangle to be isosceles, it would suffice for /DBG to be 75o. Since DB | AB that means that we need only show that /ABG is a 15o angle. Let us focus in on the top of the figure and draw in line segment HJ.
The isosceles 60o triangles, AGH, AGI, and BJK are all congruent equilateral triangles
where the length of each side is the length of the side of the regular dodecahedron. It follows that HG is both congruent and parallel to BJ. That makes quadrilateral HJBG a parallelogram since we have a theorem that a quadrilateral where one pair of opposite sides is both congruent and parallel, the figure is a parallelogram.. Since HI and IJ are both sides of the regular dodecagon, triangle HIJ is an isosceles triangle whose vertex angle is 150o. That makes /IJH a 15o angle. Since HJBG is a parallelogram, HJ || BG, and /ABG is a 15o angle as required. QED.