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.

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 $a > 0$.
