Construction of a regular pentagon 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).
 A: Here is my explanation:
Let the radius of the  circle drawn in  step 1) be $2$ (units). Then the radius of the larger 
 arc (through $A$ and $C$) drawn in step 8) is $1+\sqrt 5$ and the radius of the smaller   arc is $ \sqrt5-1$ . 
Triangle $\color{darkgreen}{\triangle hfD}$, thus, has side lengths $2$, $2$, and $\sqrt 5-1$; and so, is a 
 golden triangle (see below) with angles $36^\circ$-$72^\circ$-$72^\circ$.
Triangle $\color{red}{\triangle fhC}$ has side lengths $2$, $2$, and $\sqrt 5+1$; and thus, is a  golden triangle  with angles 
 $36^\circ$-$108^\circ$-$36^\circ$. 
From this, one can deduce that the angles $\angle ChD$ and $\angle ChB$ have measures $72^\circ$ (note the angle formed by $\ell_1$ and the segment $\overline{hC}$ is $18^\circ$).
By symmetry, the angles $\angle BhA$ and $\angle AhE$ have measure $72^\circ$. The remaining angle, $\angle EhD$ must then  have measure $72^\circ$.


On golden triangles:
The golden ratio is $\tau={1+\sqrt 5\over 2}$; 
a golden triangle is an isosceles triangle with two sides in the golden ratio.
By considering similar triangles  in the diagrams below, one can show that the golden triangles are the 
$36^\circ$-$72^\circ$-$72^\circ$ and the 
$36^\circ$-$108^\circ$-$36^\circ$ triangles.


