GF is the radical axis

Question

Given $\Delta ABC$. $D,E$ lie in $AC,AB$. $BD \cap CE =$ {$F$}. $(ADE) \cap (ABC) = ${$G,A$}. Let $M,N,P,Q$ be the midpoint of $BC,ED,EB,CD$, respectively. Prove that $G$F is the radical axis of $(GMN)$ and $(GPQ)$

Answer 1

I've translated your problem into the context of the so-called complete quadrangle. The geometry of this configuration has been studied for centuries, and is associated with such gems as the nine-point conic. There is a website devoted to it and its landmark points, lines, circles, and conics - the ENCYCLOPEDIA OF QUADRI-FIGURES (EQF) , also available as a PDF.

In the figure above, the vertices have been relabeled. The base quadrangle vertices are $A,B,C,D$, and opposite sides intersect at $X_2,X_3.$ Your original triangle $ABC$ becomes $X_2BC$, and circles $(GMN)$ and $(GPQ)$ become $(X_3M_{BC}M_{AD})$ and $(X_2M_{BD}M_{AC})$. Note that each circle corresponds to a pair of opposite sides of the quadrangle and their intersection and midpoints. By symmetry, there is a third circle $(X_1M_{DC}M_{BA})$. Because there are three pairs of opposite sides and their corresponding circles, the assertion to be proven -- that $X_1$ is on the radical axis $R_{23}$ of $(X_3M_{BC}M_{AD})$ and $(X_2M_{BD}M_{AC})$ -- has similar statements for the other $X_i$.

Although much of the geometry of the complete quadrangle can be developed with synthetic methods, it appears that the power tools of choice are analytic. One approach, for example, is to use barycentric coordinates, and to express points and loci in this form. Although there is rich geometry in OP's question, a synthetic answer has eluded me, and so we turn to barycentric coordinates.

Using methods described in the aforementioned EQF we choose the triangle $ABC$ as a base, so that $A,B,C$ have coordinates $(1:0:0),(0:1:0),(0:0:1)$ respectively, and we designate the coordinates of the fourth point $D$ as $(p:q:r)$ and the sidelengths of the base triangle as $a,b,c$. Then we build up the lines and points of the diagram. For example, the lines $AC$ and $BD$ are represented by $y=0$ and $-r x+p z=0$. And $X_1, X_2, X_3$ are $(-p:-q:0),(-p:0:-r),(0:-q:-r)$. (The calculations can be automated using a Mathematica notebook for barycentric coordinate calculations available here).

Cutting to the chase, the equation of the radical axis $R_{23}$ is

$$ \begin{align} q (-c^2 p q+a^2 p r-c^2 p r+b^2 q r-c^2 q r+a^2 r^2+b^2 r^2-c^2 r^2) &x \\ + p (c^2 p q-a^2 p r+c^2 p r-b^2 q r+c^2 q r-a^2 r^2-b^2 r^2+c^2 r^2)& y \\ -(p+q+r) (a^2 p q-b^2 p q-b^2 p r+a^2 q r) &z =0\end{align} $$

Although this is a formidable equation, it is easy to see that the coefficients for $x$ and $y$ have a common factor, and that $X_1=(-p:-q:0)$ is on this line.

With acknowledgements to the EQF community for their assistance.

Answer 2

A synthetic proof can be found in Michal Rolinek and Le Anh Dung, The Miquel Points, Pseudocircumcenter, and Euler-Poncelet Point of a Complete Quadrilateral, Forum Geometricorum, 14 (2014) 145--153. Available at http://forumgeom.fau.edu/FG2014volume14/FG201413.pdf.

See the proof for Theorem 10. Figure 7 corresponds to the diagrams given by OP and in answer https://math.stackexchange.com/a/2753547/1257, although it is flipped left to right.

It is generally frowned upon to answer questions with links, but the given citation should be quite stable, and the proof is too lengthy to reproduce here.