TITLE: Find $\frac{|AE|}{|EB|}$ in the following figure containing square and circular arc
QUESTION [5 upvotes]: $ABCD$ 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?

REPLY [2 votes]: 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
$$\frac{|AE|}{|EB|}=\frac{1-a}{a}=2.854292$$