In Robert Dixon's Mathographics, a regular pentagon is constructed with straightedge and compass only. It is the pentagon $ABCDE$ pictured below.
I am having trouble seeing why the central angles are all $72^\circ$, though. Can anyone provide the proof?
Also, does anyone know who this construction is due to? I haven't seen it anywhere, other than in Dixon's book; is it Dixon's result?
The result appears much more impressive without all the labels (which are slightly misplaced, please excuse this); however, I provided those so that answering would be easier. Also, it makes it easy to describe the steps in the construction:
1) Draw a circle (the red one) with center $h$.
2) Draw the perpendicular lines $\ell_1$ and $\ell_2$ through $h$. Locate the points of intersection $f$, $B$, and $g$ with the red circle.
3) Bisect the line segment $gh$. Denote the center by $a$.
4) Draw the green circle with center $a$ and radius $ah$.
5) Draw the other green circle (as in (3) and 4)).
6) Draw the line segment through $f$ and $a$.
7) Locate the points of intersection $b$ and $c$ of the line segment with the circle constructed in step 4).
8) Draw the blue arcs (both have center at $f$ and the radii are $fb$ and $fc$).
9) Locate the points of intersection $A$, $C$, $D$, and $E$.
I actually have the solution to the first question, and will post it unless a more elegant explanation is provided (which is probably likely). However, I find this construction particularly beautiful, and would like to know who it is attributed to (Dixon doesn't say, explicitly).