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.
 A: 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
$$\frac{-a^2-1+\sqrt{2a^2(a^2+1)}}{-(a+1)(a^2+1)}\lt \frac{1-\sqrt 2}{a+1}.$$
A: Edit: Adding the comment by the OP.
Let N be the midpoint of AC and M is a point somewhere between N and C. 

The picture clearly shows $EF > AC$ because $EF = UV + VC = UV + AC$.
Thus, we only need to study the case when M is somewhere between A and N. See below. 


The picture shows the following idea:- As $E$ slides along the $y$-axis (and $F$ moves correspondingly), there is a critical position $E’$ ($F’$ correspondingly) such that $AC = E’F’$. If $AE < AE’$, $EF$ will be shorter than $AC$.
To make the two length comparable, I translate $E’F’$ to $XC$ such that $E'F'CX$ is a //gm. $CX$ is extended to cut the $y$-axis at $Y$. The circle drawn with $C$ as center and $CA$ as radius shows more clearly of the effect.
$E’A = E’B – AB = XZ – 1 = \sqrt(2) \cos \beta – 1$; where $\beta$ is the angle that EF inclines to the y-axis.
A: In my solution to 
the original problem,
I used algebra and 
analytic geometry.
I tried to do the same thing
for this generalization,
but the final simplification
does not occur here.
Here is what I've got.
Suppose
$AB = BC = 1$.
Then
$M
=(a, 1-a)
$,
where
$0 < a < 1$.
Let
$E = (0, 1+v)
$,
where $v > 0
$.
Then
line $EM$
is
$\frac{y-(1+v)}{x}
=\frac{(1-a)-(1+v)}{a}
=\frac{-a-v}{a}
=-1-\frac{v}{a}
$
or
$y
=1+v-x(1+\frac{v}{a})
$.
Putting
$y=0$,
$x
=\frac{1+v}{1+\frac{v}{a}}
$,
so
$F
=(\frac{1+v}{1+\frac{v}{a}}, 0)
$.
$|AC|^2
=2
$
$\begin{array}\\
|EF|^2
&=(\frac{1+v}{1+v/a})^2+(1+v)^2\\
&=(1+v)^2(1+\frac{1}{(1+v/a)^2})\\
&=(1+v)^2(\frac{1+(1+v/a)^2}{(1+v/a)^2})\\
&=(1+v)^2(\frac{a^2+(a+v)^2}{(a+v)^2})\\
&=(1+v)^2(\frac{2a^2+2av+v^2}{(a+v)^2})\\
\end{array}
$
According to Wolfy,
$\begin{array}\\
|EF|^2-|AC|^2
&=(1+v)^2 (\frac{2a^2+2av+v^2}{(a+v)^2}) -2\\
&=v \frac{2 a^2 v+4 a^2+2 a v^2+4 a v-2 a+v^3+2 v^2-v}{(a+v)^2}\\
\end{array}
$
Therefore, 
when
$0
\lt 2 a^2 v+4 a^2+2 a v^2+4 a v-2 a+v^3+2 v^2-v\\
= a^2(2v+4)+a(2v^2+4v-2)+v^3+2 v^2-v\\
$
for
$0 < a < 1$
and
$v > 0
$,
|EM| > |AC|.
Again
according to Wolfy,
if
$a = 1-c$,
then
$f(v, a)\\
=2 a^2 v+4 a^2+2 a v^2+4 a v-2 a+v^3+2 v^2-v\\
=2 c^2 (v+2)-2 c (v^2+3 v+1)+v^3+3 v^2+3 v/2\\
$
If $c \le 0$
(i.e., $a \ge \frac12$),
all the terms are positive,
so |EM| > |AC| for all $v > 0$.
However,
if $c > 0$
(i.e., $a < \frac12$),
for small $v$
this is about
$4c^2-2c
=2c(2c-1)
\lt 0
$,
so
there are values of $v$
for which the expression is negative
so that
|EM| < |AC|.
Again according to Wolfy,
this occurs at the following values of $v$:
$
a=1/3: v< 0.19795\\
a=1/4: v< 0.27832\\
a=1/10: v< 0.38509\\
a=1/100: v< 0.41382\\
a=1/10000: v<0.41421\\
$
Looking at these values,
it seems that
the $v$ bound is approaching
$\sqrt{2}-1$.
To show this is true,
Wolfy gets
$f(\sqrt{2}-1, a)
=2(1+\sqrt{2})a^2
\gt 0
$
for $a > 0$.
