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

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.

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.