<s>[INST] Find BC/AB if I lies in the circumcircle of AEF Let $ABC$ be a right triangle with angle $B=\frac{\pi}{2}$. Let $E$ and $F$ be the midpoint of $AB$ and $AC$ respectively. If $I$ the in centre of $ABC$ lies on the circumcircle of $AEF$, find the ratio $BC/AB$. 
My guess would be that $AB=BC$ but I am unable to prove it. I tried angle chasing but that doesn't seem to lead me anywhere. Or is this possible by using coordinates?
 [/INST] Without loss of generality you may choose coordinates
$$A=\begin{pmatrix}0\\0\end{pmatrix}\qquad
B=\begin{pmatrix}1\\0\end{pmatrix}\qquad
C=\begin{pmatrix}1\\\lambda\end{pmatrix}\qquad
E=\begin{pmatrix}1/2\\0\end{pmatrix}\qquad
F=\begin{pmatrix}1/2\\\lambda/2\end{pmatrix}$$
Now the barycentric coordinates of the incenter are $a:b:c$ and we have $a=\lambda$, $b=\sqrt{1+\lambda^2}$ and $c=1$. So the incenter has coordinates
$$I = \frac{aA+bB+cC}{a+b+c}=
\frac{1}{\lambda+\sqrt{1+\lambda^2}+1}\begin{pmatrix}
\sqrt{1+\lambda^2}+1\\\lambda
\end{pmatrix}$$
Now you want $A,E,F,I$ cocircular. Looking at this answer of mine, you can check that by computing the determinant
$$\begin{vmatrix}
x_1^2+y_1^2 & x_1 & y_1 & 1 \\
x_2^2+y_2^2 & x_2 & y_2 & 1 \\
x_3^2+y_3^2 & x_3 & y_3 & 1 \\
x_4^2+y_4^2 & x_4 & y_4 & 1
\end{vmatrix}=0$$
So in this case:
$$\begin{vmatrix}
0 & 0 & 0 & 1 \\
1/4 & 1/2 & 0 & 1 \\
1/4+\lambda^2/4 & 1/2 & \lambda/2 & 1 \\
\frac{\left(\sqrt{1+\lambda^2}+1\right)^2+\lambda^2}
{\left(\lambda+\sqrt{1+\lambda^2}+1\right)^2} &
\frac{\sqrt{1+\lambda^2}+1}{\lambda+\sqrt{1+\lambda^2}+1} &
\frac{\lambda}{\lambda+\sqrt{1+\lambda^2}+1} &
1
\end{vmatrix}=0$$
Developing the determinant by the first row simply means restricting things to the lower left $3\times3$ determinant. You can further simplify the equation by multiplying rows by their denominator.
$$\begin{vmatrix}
1 & 2 & 0 \\
1+\lambda^2 & 2 & 2\lambda \\
\left(\sqrt{1+\lambda^2}+1\right)^2+\lambda^2 &
\left(\sqrt{1+\lambda^2}+1\right)
\left(\lambda+\sqrt{1+\lambda^2}+1\right) &
\lambda\left(\lambda+\sqrt{1+\lambda^2}+1\right)
\end{vmatrix}=0$$
Unfortunately the square roots haven't canceled yet, so I'll ask a computer algebra system for help here.
sage: PR1.<x,b> = QQ[]
sage: m = matrix([
...     [1,2,0],
...     [1+x^2,2,2*x],
...     [(b+1)^2+x^2,(b+1)*(x+b+1),x*(x+b+1)]])
sage: d = m.det()
sage: ideal([d, x^2 + 1 - b^2]).variety(QQbar)
[{b: -1, x: 0},
 {b: 0, x: -1*I},
 {b: 0, x: 1*I},
 {b: 1, x: 0},
 {b: 1.666666666666667?, x: 1.333333333333334?}]

So there are five algebraic solutions. Two have a complex value for $\lambda$, leading to a zero length $b$ which is not geometrically sensible since it would imply $A=C$. (Speaking about lengths and points at complex coordinates in the same setup would be tricky business, and probably should start with a clear definition of the terms involved.)
Two have $\lambda=0$, so $B=C$, in which case many of the objects involved are not well defined.
The only remaining solution is for $\lambda=\frac43$ as Lucian wrote.
</s>