<s>[INST] Minimizing lengths of cevians in an isosceles right triangle Consider isosceles right triangle $ABC$ with $BC$ as the hypotenuse and $AB=AC=6$. $E$ is on $BC$ and $F$ is on $AB$ such that $AE+EF+FC$ is minimized. Compute $EF$.
My thought process:
I reflected triangle $ABC$ across $BC$ to get a square $ABA'C$. Then, I messed around with the placement of $F$ on $AB$, to see what results I could produce, with $E$ always being the midpoint. May someone help me on this?
 [/INST] 
Let $|AB|=|AC|=6=b$.
It is known that the minimal path would be 
the path of the light ray from the point $C$
to point $A$, reflected at points $F$ and $E$.
Consider points $C_1,A_1$ and $E_1$
as reflection of the points $C,A$ and $E$
with respect to $AB$.
Let 
$\angle A_1CC_1=\angle CFG=\angle GFE=\phi$,
$\angle FEH=\angle HEA=\angle AE_1H=\angle HE_1C=\theta$.
Then
\begin{align}
|EF|&=|E_1F|,\quad |EA|=|E_1A|=|E_1A_1|
,\\
|AE|+|EF|+|FC|&=
|A_1E_1|+|E_1F|+|FC|
\\
&=|CA_1|
=\sqrt{(2b)^2+b^2}=\sqrt5\,b
.
\end{align} 
\begin{align} 
\triangle BEF:\quad
\angle BEF=90^\circ-\theta
,\quad 
\angle EFB=90^\circ-\phi
\quad\Rightarrow\quad
\theta=45^\circ-\phi
,
\end{align}
\begin{align} 
\triangle A_1CC_1:\quad
\phi&=\arctan\tfrac12
=\arcsin\tfrac{\sqrt5}5
=\arccos\tfrac{2\sqrt5}5
,\\
\theta&=45^\circ-\arcsin\tfrac{\sqrt5}5
=45^\circ-\arccos\tfrac{2\sqrt5}5
. 
\end{align} 
\begin{align} 
\triangle BEF:\quad
\frac{|EF|}{\sin45^\circ}
&=
\frac{|BF|}{\sin(90^\circ-\theta)}
=
\frac{\tfrac12|AB|}{\cos\theta}
,\\
|EF|&=\frac{b\sqrt2}{4\cos(45^\circ-\arctan\tfrac12)}
\\
&=
\tfrac{\sqrt2}4\,
\frac{b}{
\cos 45^\circ \cos(\arccos\tfrac{2\sqrt5}5)
+
\sin 45^\circ \sin(\arcsin\tfrac{\sqrt5}5)
}
\\
&=
\tfrac{\sqrt2}4\,
\frac{b}{
\tfrac{\sqrt2}2 
\tfrac{2\sqrt5}5
+
\tfrac{\sqrt2}2
\tfrac{\sqrt5}5
}=b\,\tfrac{\sqrt5}6
,
\end{align}
and since $b=6$,
the answer is $|EF|=\sqrt5$.
</s>