Find $\frac{|AE|}{|EB|}$ in the following figure containing square and circular arc $ABCD$ One frame and $E\in[AB]$. The point where $[DE]$ cuts the arc of the circle with center $B$ and radius $[AB]$ is $F$, if $|FE|=|FC|$ is $\frac{|AE|}{|EB|} =?$

To make the question easier, I accepted one side of the square as a unit. Then I tried to decipher it by typing the coordinates. $A(0,0),B(1,0),C(1,1),D(0,1),E(m,0), d_{DE}:y=\frac{-x}{m}+1,\ \ \ \bigcirc : (x-1)^2+y^2=1$ Wrote. Now $d_{DE}\cap\bigcirc = \{F\}$ I used to reach the coordinates of point $F$. But it's kind of hard from now on. $|FE|=|FC|$ We can use it to get results but there's a wall in front of me: $m^6-4m^5+5m^4-8m^3+8m^2-8m+4=0$ Can you help me solve the equation or a solution with basic geometry?
 A: If use the polar coordinates in the circle of radius 1, the coordinates of the points are: $E(a,0)$, $F(\cos{\theta}), \sin{\theta}$, $A(1,0)$,  $B(0,0)$, $C(0,1)$, $D(1,1)$ and define the variables
$$\sin{\theta}=u, \cos{\theta=v}\rightarrow u^2+v^2=1,$$
I found the folowing equations:
a) condition that $D,E,F$ be collinears $\rightarrow$
$$a=\frac{u-v}{u-1}$$
b) condition that $|FE|=|FC|\rightarrow$
$$a^2-2va+2u-1=0\rightarrow$$
Replacing $a$ in the last eq.:
$$u^2(3+2v)-6u+2=0$$
with the real root $u=0.661588\rightarrow a= 0.259451$ (Wolfram|Aalpha).
Hence
$$\frac{|AE|}{|EB|}=\frac{1-a}{a}=2.854292$$
A: The equation of the line with endpoints E and D is
$y = (\frac{1}{m})x + 1 $
This intercepts another linear line at $(x,y)$ which has the equation:
$y = (\frac{x - 1}{y - 1})x + \frac{y - x}{y - 1}  \implies y=1+\sqrt{x\left(x-1\right)}$
So through substitution $(x,y)\implies (\frac{m^2}{m^2-1},\frac{m^2 + m  - 1}{m^2 - 1})$
Now note that
$|AB| = 1 - \frac{m^2 + m - 1}{m^3 - m}  + \frac{m^2}{m^2-1} + |EB| \implies |EB| = \frac{2m - 1}{m}$
Which also gives $|AE| = \frac{m - 1}{m}$
Now we can determine another equation for $|AE|$ quite easily to solve for $m$(using Pythagorean theorem and $|DE|$) to find the ratio of $\frac{|AE|}{|EB|}$ which is in the form $\frac{m -1}{2m - 1}$.
A: 
Hints: Use this figure and try to find relations.The center  O of circumcircle of triangle FEC is on altitude FH.
$\frac {AE}{EB}\approx 3$
This value must satisfy the equation you found.
