Proof of Pentagon Construction

I was walking through the following construction of a regular pentagon from the AoPS Geometry textbook recently.

• Quarter a circle
• Construct the midpoint of one radius at F.
• Create triangle AFO 
• Bisect angle AFO  with segment FG
• Construct a perpendicular BG at the point G from the diameter of the circle to the circumference.
• Segment AB is $ \frac{1}{5} $ of the circumference and you can now copy it around the circle to form the pentagon.

But why does this work?   What follows is a cleaned up explanation with some improvements borrowed from http://www.cut-the-knot.org

First the pentagon divides the circumference of the circle into five 72 degree arcs so it would be nice to understand the side length ratios of the 18-72-90 triangle.  To get there we’ll start with its close cousin the 36-72-72. (See: Deep Dive)

• We’re going to bisect angle A with the segment AC. This forms 2 more isosceles triangle ABC and ACO and all the segments BC, AC and AO are congruent.  Moreover triangle ACO is another smaller 36-72-72
• So let  AO = 1 and let CO = a
• From the similar triangles ACO and ABO  we get $ \frac{1}{a} = \frac{a+1}{1} $ 
• Simplifying $ a^2 + a - 1 = 0 $  and $ a = \frac{-1 + \sqrt{5}}{2} $.  Hey that’s almost the golden ratio. In fact $a  = \phi - 1 $  and the entire side BO = a + 1 = $ \phi $  
• We can then add a perpendicular bisector to derive the following ratios for the 18-72-90

Now lets return to the construction. Without loss of generality let’s set the radius of the circle to also be the golden ratio $ \phi $.   This will make the comparisons easier. It now remains to show segment OG is length of $ \frac{1}{2} $  If this is the case then triangle BOG is congruent to the one above and angle BOG is 72 degrees.  From there the arc AB is $ \frac{1}{5} $ of the circle and we have our first pentagon side.

• Since the circle is radius $ \phi $, BO is already identical to the 18-72-90 triangle hypotenuse.
• Looking at triangle AFO, FO is $\frac{\phi}{2}$, AO is $ \phi $ so via the Pythagorean Theorem AF is $ \sqrt{5} \cdot  \frac{\phi}{2} $
• We can now use the angle bisector theorem to find OG.

\[\frac{OG}{FO} = \frac{AG}{AF}\] \[OG = \frac{\phi}{2} \cdot \frac{ \phi - OG}{\sqrt{5} \cdot \frac{\phi}{2}} \]

That simplifies to   $ (\sqrt{5} + 1)OG = \phi $ 

but  since $ (\sqrt{5} + 1) = 2 \phi $ we get $ 2 \phi \cdot OG = \phi $ and $ OG = \frac{1}{2} $ as needed!