A novel (?) construction of the regular pentagon with straightedge and compass

Question

With reference to the triangle $\triangle ABC$ illustrated in the picture below, given the side $AC$, the five points $B,D,E,F,G$, in the conditions discussed here, determine a circle (red).

Let us consider the case in which $\triangle ABC$ is isosceles. In this case, the point $B$ lies on the perpendicular bisector (dashed line) of the side $\overline{AC}$.

Now, we draw the circle with center in $C$ and passing through $A$ (green) and the prolongation of the side $BC$ (brown), obtaining the points $H$ and $I$.

Since $B$ must lie on the dashed line, there is only one case in which the points $H$ and $I$ coincide:

My conjecture is that, if $H\equiv I$, the points $B,D,E,F,G$ determine a regular pentagon.

I wonder if you can help me to prove or disprove such conjecture.

Thanks for your help! I apologize in case of incorrectness or trivialities.

Answer 1

In all pics take $a=AF=AD=CE$, $b=BF=DE$. Thus the sides of $ABC$ are $a+b$ resp. $2a+b$. Further you have $AI=a+b$ and $CH=2a+b$.

Thus, when $H=I$ as for the last 2 pics, the triangle $AC(H=I)$ likewise has just those side lengths. Therefore you have the pair of golden triangles, the obtuse and the acute one here. And from that it follows that the pentagon $BFDEG$ truely is regular, having side length $b$.

You then even could deduce that $D$, $F$, and $(H=I)$ are collinear and that $F(H=I)=a$ as well.

--- rk