Geometry: prove that point M is touched by 4 circles

[Jiketz]

Homework Statement

This is the problem: Given are two parallelograms ABCD and AECF with common diagonal
diagonal AC, where E and F lie inside the parallelogram ABCD.
Show:
The circumcircles of the triangles AEB, BFC, CED and DFA have a point
in common.
I've already given an answer in the pdf, but how can I prove that the point M is also on the remaining two circles? Here is another sketch I drew. Can you please finish the pdf paper ?

Relevant Equations

relevant equations can be found in the pdf solution from me

 

[anuttarasammyak]

Interesting exercise. I take Cartesian coordinates to look at as attached figure.
We can write the equation of circle touching three points as
$$x^2+y^2+lx+my+n=0$$
Giving three (x,y)s of touching points, we get a set (l,m,n) . We get four (l,m,n) sets for the four circles. The system of four equations of the circles would show one degenerate solution.
That's the design of the proof, though I have not tired yet due to an expected terrible matrix calculation. The symmetry should make it easy.

 

[Lnewqban]

Welcome, @Jiketz !

The relations shown at this link may help:
https://en.wikipedia.org/wiki/Circumscribed_circle

 

[anuttarasammyak]

Contd. from my previous post

The system of four equations of the circles would show one degenerate solution.

The equation of line which connects two commom points of the cirlces i and j, is
$$(l_i-l_j)x+(m_i-m_j)y+n_i-n_j=0$$
The system of such six lines meet at a point (x,y) are recuced to three equations.
$$\begin{pmatrix} l_1-l_2 & m_1-m_2 & n_1-n_2 \\ l_2-l_3 & m_2-m_3 & n_2-n_3 \\ l_3-l_4 & m_3-m_4 & n_3-n_4 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ 1 \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix}$$
This 3X3 matrix in LHS shoud have determinant zero so that there is a solution.
$$\begin{vmatrix} l_1-l_2 & m_1-m_2 & n_1-n_2 \\ l_2-l_3 & m_2-m_3 & n_2-n_3 \\ l_3-l_4 & m_3-m_4 & n_3-n_4 \\ \end{vmatrix} =\sum_{i,j,k,q=1}^4 P(ijkq)l_i m_j n_k$$
where P(ijkq)=0 when any of them equals.
P(ijkq)=1 when ijkq is even pertumation of 1234.
P(ijkq)=-1 when ijkq is odd permutation of 1234.

We expect to get coordinates of M as
$$\begin{pmatrix} l_1-l_4 & m_1-m_4 & n_1-n_4 \\ l_2-l_4 & m_2-m_4 & n_2-n_4 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ 1 \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \end{pmatrix}$$
or
$$\begin{pmatrix} x \\ y \\ \end{pmatrix} =\frac{-1}{(l_1-l_4 )(m_2-m_4 )-(l_2-l_4 )(m_1-m_4 )} \begin{pmatrix} m_2-m_4 & m_4-m_1 \\ l_4-l_2 & l_1-l_4 \\ \end{pmatrix} \begin{pmatrix} n_1-n_4 \\ n_2-n_4 \\ \end{pmatrix}$$

 

[anuttarasammyak]

Supplement to my post #2

Say three points $(a_1,a_2),(b_1,b_2),(c_1,c_2)$ is on the circle which is caratterized by (l,m, n)
$$\begin{pmatrix} a_1& a_2 & 1 \\ b_1& b_2 & 1 \\ c_1& c_2 & 1 \\ \end{pmatrix} \begin{pmatrix} l \\ m \\ n \\ \end{pmatrix} =- \begin{pmatrix} a_1^2+a_2^2 \\ b_1^2+b_2^2 \\ c_1^2+c_2^2 \\ \end{pmatrix}$$
Solving it for l,m,n
$$\begin{pmatrix} l \\ m \\ n \\ \end{pmatrix} =\frac{-1}{det} \begin{pmatrix} b_2-c_2 &　c_2-a_2 &　a_2-b_2　 \\ c_1-b_1 & a_1-c_1 & b_1-a_1　\\ b_1c_2-b_2c_1　 & c_1a_2-c_2a_1　& a_1b_2-a_2b_1 \\ \end{pmatrix} \begin{pmatrix} a_1^2+a_2^2 \\ b_1^2+b_2^2 \\ c_1^2+c_2^2 \\ \end{pmatrix}$$
where
$$det=a_1b_2-a_2b_1+b_1c_2-b_2c_1+c_1a_2-c_2a_1$$

Parameters in this exercise are
$$\begin{matrix} & CircleDAF & CircleDCE & CircleBCF & CircleBAE \\ a_1 & d_1 & d_1 & -d_1 & -d_1 \\ a_2 & d_2 & d_2 & -d_2 & -d_2 \\ b_1 & f_1 & -f_1 & f_1 & -f_1 \\ b_2 & f_2 & -f_2 & f_2 & -f_2 \\c_1 & -1 & 1 & 1 & -1 \\ c_2 & 0 & 0 & 0 & 0 \\ det & -d_2+f_2+d_1f_2-d_2f_1 & d_2+f_2-d_1f_2+d_2f_1 & -d_2-f_2-d_1f_2+d_2f_1 & d_2-f_2+d_1f_2-d_2f_1 \end{matrix}$$
As for circle DAF
$$\begin{pmatrix} l \\ m \\ n \\ \end{pmatrix} =\frac{-1}{det} \begin{pmatrix} f_2 & -d_2 &　d_2-f_2　\\ -1-f_1 & d_1+1 & -d_1+f_1 \\ f_2& -d_2 & d_1f_2-d_2f_1 \\ \end{pmatrix} \begin{pmatrix} d_1^2+d_2^2 \\ f_1^2+f_2^2 \\ 1 \\ \end{pmatrix}$$
$$=\frac{-1}{det} \begin{pmatrix} f_2 |d|^2 -d_2 |f|^2 +d_2-f_2　\\ (-1-f_1 )|d|^2+ (d_1+1 )|f|^2 -d_1+f_1 \\ f_2|d|^2 -d_2|f|^2+d_1f_2-d_2f_1 \\ \end{pmatrix}$$
where
$$|d|^2=d_1^2+d_2^2$$
$$|f|^2=f_1^2+f_2^2$$
$$det=-d_2+f_2+d_1f_2-d_2f_1$$

As for circle DCE
$$\begin{pmatrix} l \\ m \\ n \\ \end{pmatrix} =\frac{-1}{det} \begin{pmatrix} -f_2 & -d_2 & d_2+f_2　\\ 1+f_1 & d_1-1 & -d_1-f_1\\ f_2 & d_2 & -d_1f_2+d_2f_1 \\ \end{pmatrix} \begin{pmatrix} d_1^2+d_2^2 \\ f_1^2+f_2^2 \\ 1 \\ \end{pmatrix}$$
$$=\frac{-1}{det} \begin{pmatrix} -f_2 |d|^2 -d_2 |f|^2+ d_2+f_2　\\ (1+f_1) |d|^2+ (d_1-1 )|f|^2 -d_1-f_1\\ f_2 |d|^2+ d_2|f|^2 -d_1f_2+d_2f_1 \\ \end{pmatrix}$$
where
$$det=d_2+f_2-d_1f_2+d_2f_1$$

As for circle BCF
$$\begin{pmatrix} l \\ m \\ n \\ \end{pmatrix} =\frac{-1}{det} \begin{pmatrix} f_2 & d_2& -d_2-f_2\\ 1-f_1 & -1-d_1 & d_1+f_1\\ -f_2 & -d_2 & -d_1f_2+d_2f_1 \\ \end{pmatrix} \begin{pmatrix} d_1^2+d_2^2 \\ f_1^2+f_2^2 \\ 1 \\ \end{pmatrix}$$
$$=\frac{-1}{det} \begin{pmatrix} f_2 |d|^2+ d_2|f|^2 -d_2-f_2\\ (1-f_1)|d|^2 +( -1-d_1) |f|^2+ d_1+f_1\\ -f_2 |d|^2 -d_2 |f|^2 -d_1f_2+d_2f_1 \\ \end{pmatrix}$$
where
$$det=-d_2-f_2-d_1f_2+d_2f_1$$
for circle BAE

$$\begin{pmatrix} l \\ m \\ n \\ \end{pmatrix} =\frac{-1}{det} \begin{pmatrix} -f_2 & d_2 & -d_2+f_2 \\ -1+f_1 & -d_1+1 & d_1-f_1 \\ -f_2 & d_2 & d_1f_2-d_2f_1 \\ \end{pmatrix} \begin{pmatrix} d_1^2+d_2^2 \\ f_1^2+f_2^2 \\ 1 \\ \end{pmatrix}$$
$$=\frac{-1}{det} \begin{pmatrix} -f_2 |d|^2+ d_2 |f|^2 -d_2+f_2 \\ (-1+f_1)|d|^2+ (-d_1+1 )|f|^2+ d_1-f_1 \\ -f_2 |d|^2+ d_2 |f|^2+ d_1f_2-d_2f_1 \\ \end{pmatrix}$$
where
$$det=d_2-f_2+d_1f_2-d_2f_1$$
I hope I have not made careless mistake.