Problem: Let $ABC$ be a triangle with incenter $I$
and let $D$ be an arbitrary point on the side $BC$.
Let the line through $D$ perpendicular to $BI$ intersect $CI$ at $E$.
Let the line through $D$ perpendicular to $CI$ intersect $BI$ at $F$.
Prove that the reflection of $A$ across the line $EF$ lies on the line $BC$.

Solution: We begin as follows:
\begin{claim*}
  $AEIF$ is cyclic.
\end{claim*}
\begin{proof}
  Let $X = \ol{CA} \cap \ol{DF}$.
  Then
  \begin{align*}
    \dang FXA &= \dang DXC = \dang CDX = 90\dg - \half C
    = \dang AIB = \dang FIA \\
    \dang EXF &= \dang EXD = \dang XDE = \dang FDE = \dang EIF
  \end{align*}
  which implies $AFXI$ and $FXIE$ are cyclic, respectively.
\end{proof}

\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair D = 0.7*C+0.3*B;
pair I = incenter(A, B, C);
pair E = extension(C, I, D, foot(D, B, I));
pair F = extension(B, I, D, foot(D, C, I));

filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, deepcyan);
draw(B--F, lightblue);
draw(C--I--A, lightblue);
filldraw(D--E--F--cycle, opacity(0.1)+lightred, red);

filldraw(circumcircle(A, E, F), opacity(0.1)+yellow, deepgreen+dashed);
pair Y = extension(D, E, A, B);
pair X = extension(D, F, A, C);
draw(D--Y, red);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(270));
dot("$I$", I, dir(250));
dot("$E$", E, dir(225));
dot("$F$", F, dir(F));
dot(Y);
dot("$X$", X, dir(X));

/* TSQ Source:

A = dir 110
B = dir 210
C = dir 330
D = 0.7*C+0.3*B R270
I = incenter A B C R250
E = extension C I D foot D B I R225
F = extension B I D foot D C I

A--B--C--cycle 0.1 lightcyan / deepcyan
B--F lightblue
C--I--A lightblue
D--E--F--cycle 0.1 lightred / red

circumcircle A E F 0.1 yellow / deepgreen dashed
Y .= extension D E A B
X = extension D F A C
D--Y red

*/
\end{asy}
\end{center}
Now, the dilation of the Simson line of $A$
to $\triangle IEF$ should be collinear,
but the reflections of $A$ about lines $BI$ and $CI$
lie on line $BC$ by definition.
This solves the problem.