When is EF longer than AC? (a generalization) ABC is an isosceles right triangle, 
M is on AC, 
and EMF is a straight line.
When is EF longer than AC?
]1
Note:
  This is a generalization
of the following problem,
which has M in the
center of AC:
Prove that EF is longer than AC
In that case,
I showed that
EF is always longer than AC.
I can show the following:
If M is closer to C than A,
then EF is longer.
If M is closer to A than C,
then EF can be shorter,
but it will be longer when
EA is long enough.
If $|EB| \ge \sqrt{2}$,
then EF is always longer
no matter where M is.
If there are no answers in two days,
I will post mine.

We may suppose that $$A(0,1),B(0,0),C(1,0),M(m,1-m),E(0,1-m-am),F\left(m+\frac{m-1}{a},0\right)$$
where $0\lt m\lt 1$ is the $x$ coordinate of $M$ and $a\lt -1$ is the slope of the line $EF$.
Then, we have
$$\begin{align}&|EF|\gt |AC|\\\\&\iff |EF|^2\gt |AC|^2\\\\&\iff \left(m+\frac{m-1}{a}\right)^2+(1-m-am)^2\gt 2\\\\&\iff m^2+\frac{2m(m-1)}{a}+\frac{(m-1)^2}{a^2}+1+m^2+a^2m^2-2m-2am+2am^2-2\gt 0\\\\&\iff (a+1)^2(a^2+1)m^2-2(a+1)(a^2+1)m+(a+1)(1-a)\gt 0\\\\&\iff (a+1)(a^2+1)m^2-2(a^2+1)m+1-a\lt 0\\\\&\iff \color{red}{\frac{-a^2-1+\sqrt{2a^2(a^2+1)}}{-(a+1)(a^2+1)}\lt m\lt 1}\tag1\end{align}$$
Here note that 
$$0\lt f(a)=\frac{-a^2-1+\sqrt{2a^2(a^2+1)}}{-(a+1)(a^2+1)}\lt \color{blue}{\frac 12}$$
for $a\lt -1$ :
$$\begin{align}f(a)&=\frac{-a^2-1+\sqrt{2a^2(a^2+1)}}{-(a+1)(a^2+1)}\cdot\frac{-a^2-1-\sqrt{2a^2(a^2+1)}}{-a^2-1-\sqrt{2a^2(a^2+1)}}\\&=\frac{a-1}{-a^2-1-\sqrt{2a^2(a^2+1)}}\\&=\frac{1-\frac 1a}{-a-\frac 1a+\sqrt{2(a^2+1)}}\end{align}$$
is increasing for $a\lt -1$ with
$$\lim_{a\to -\infty}f(a)=0,\qquad\lim_{a\to -1^-}f(a)=\frac 12.$$
Added : Now I'm going to check if my answer agrees with what the OP wrote.

If M is closer to C than A, then EF is longer.

If $m\ge\frac 12$, then $(1)$ holds for any $(a,m)$.

If M is closer to A than C, then EF can be shorter, but it will be longer when EA is long enough.

If $m\lt\frac 12$, then EF can be shorter, but it will be longer when $a$ is small enough.

If $|EB|\ge \sqrt 2$, then EF is always longer no matter where M is.

If $1-m-am\ge \sqrt 2$, i.e. $m\ge\frac{1-\sqrt 2}{a+1}$, then $(1)$ holds for any $(a,m)$ because we have
$$\frac{-a^2-1+\sqrt{2a^2(a^2+1)}}{-(a+1)(a^2+1)}\lt \frac{1-\sqrt 2}{a+1}.$$
