$\def\conj#1{\overline{#1}}\def\i{\mathbf{i}}$
Let $ABC$ be an acute triangle with orthocenter $H$. Let $E,F$ be the
feet of the altitudes from $B,C$. Let the midpoints of $BF,CE$ be
$M,N$. Let $MN$ meet $EF$ at $K$, and let $L$ be the midpoint of
$MN$. Prove that $AL\perp HK$.
Notational remark:
in all complex expressions that follow,
$\i$ is used to represent an imaginary unit,
complex conjugate of the complex number $z=(x,y)=x+\i\,y$
is denoted as $\conj{z}=(x,-y)=x-\i\,y$,
$\Re(z)=x$ is the real part of $z$,
$\Im(z)=y$ is the imaginary part of $z$.
This proof is based on three results from the geometry of complex numbers
(for the reference,
see Liang-Shin Hahn. Complex Numbers and Geometry, 1994. isbn: 0883855100)
and the help of Computer Algebra System Maxima
on simplifications of algebraic expressions.
1) Two lines through points $z_1,z_2$
and $z_3,z_4$ intersect at a point
\begin{align}
z=
f_z(z_1,z_2,z_3,z_4)=
\frac{
(z_1-z_2)\,(\conj{z_3}\,z_4-\conj{z_4}\,z_3)
-
(z_3-z_4)\,(\conj{z_1}\,z_2-\conj{z_2}\,z_1)
}{
(z_1-z_2)\,(\conj{z_3}-\conj{z_4})
-
(z_3-z_4)\,(\conj{z_1}-\conj{z_2})
}
.
\end{align}
2) Given the coordinates $A,B,C$ as complex numbers,
the foot of perpendicular from the vertex $A$
onto the opposite side $BC$
can be found as
\begin{align}
f_{H_a}(A,B,C)&=
\tfrac12 A+
\frac{(\conj{A}-\conj{B})\,C+(\conj{C}-\conj{A})\,B}{2(\conj{C}-\conj{B})}
.
\end{align}
3) For four distinct points $\alpha,\beta,\gamma,\delta\in\mathbb(C)$
\begin{align}
\overrightarrow{\alpha\beta} \perp \overrightarrow{\gamma\delta}
&\Longleftrightarrow
\frac{\beta-\alpha}{\delta-\gamma} \quad\text{ is purely imaginary}
.
\end{align}
Let $A=(0,0)$ be the origin of the coordinate system,
$B=(c,0)$ and $C=(u,v)$ for $c,u,v\in\mathbb{R_+}$. Then
\begin{align}
A&=0+\i\,0,
\quad
B=c+\i\,0,
\quad
C=u+\i\,v,
\quad
F=u+\i\,0
,
\\
E&=f_{H_a}(B,C,A)=\frac{c\,u}{u-\i\,v}
,
\quad
H=f_z(C,F,B,E)
=-{{\i\,u\,\left(\i\,v+u-c\right)}\over{v}}
,
\\
M&=\tfrac12(B+F) = \tfrac12(u+c)
,
\quad
N=\tfrac12(C+E) = \tfrac12\left(\i\,v+u-{{c\,u}\over{\i\,v-u}}\right)
,
\\
L&=\tfrac12(M+N) =
-{{v^2-\i\,u\,v-\i\,c\,v+2\,u^2+2\,c\,u}\over{4\,\left(\i\,v-u\right)}}
,
\\
K&=f_z(M,N,E,F)\\
&= {{\i\,u\,v^3+\i\,c\,v^3-u^2\,v^2-2\,c\,u\,v^2+c^2\,v^2+\i\,u^3\,v-\i\,c
^2\,u\,v-u^4-c\,u^3+c^2\,u^2+c^3\,u}\over{2\,\left(\i\,v-u\right)\,
\left(v^2+u^2-c^2\right)}}
,
\\
X&=\frac{A-L}{H-K}
={
{v\,\left(v^2+u^2-c^2\right)
\left(v^2-\i\,u\,v-\i\,c\,v+2\,u^2+2\,
c\,u\right)}
\over
{2\,\left(u-c\right)\,\left(v^2+u^2-c\,u\right)\,
\left(\i\,v^2+u\,v+c\,v+2\,\i\,u^2+2\,\i\,c\,u\right)}
}
\\
&=\i\cdot{{v\,\left(v^2+u^2-c^2\right)}
\over
{2\,\left(c-u\right)\,\left(v^2 +u^2-c\,u\right)}}
.
\end{align}
That is, $(A-L)/(H-K)$ is purely imaginary,
$(A-L)=(H-K)\cdot\i\cdot\Im(X)$.
In other words,
vector $\overrightarrow{LA}$
is equal to vector $\overrightarrow{KH}$, scaled by real number $\Im(X)$
and rotated by 90 degrees counterclockwise
(provided $\Im(X)\ne0$, $\Im(X)\ne\pm\infty$).
Note that $(v^2+u^2=b^2)$
and hence when $|AC|=|AB|$,
$MN||EF$ (Fig.~3)
and the point $K$ is at infinity
since $v^2+u^2-c^2=0$.
An expression for $X$ reveals another two singular conditions,
when $u=c$ and $v^2 +u^2=c\,u$, but they correspond
to the right triangles
with $\angle B=90^{\circ}$ and $\angle C=90^{\circ}$ respectively,
and can be ignored since the $\triangle ABC$ is assumed to be acute.
To check the calculations,
here is the output of a Maxima session:
(%i2) display2d:false
(%i3) declare([c,u,v],real,[A,B,C,E,F,H,M,N,L,K,X],complex)
(%i4) fHa(A,B,C):=
((1/2)*((conjugate(A)-conjugate(B))*C
+(conjugate(C)-conjugate(A))*B))
/(conjugate(C)-conjugate(B))+(1/2)*A
(%i5) fz(P,Q,p,q):=
((P-Q)*(p*conjugate(q)-q*conjugate(p))
-(p-q)*(P*conjugate(Q)-Q*conjugate(P)))
/((Q-P)*(conjugate(p)-conjugate(q))
+(q-p)*(conjugate(Q)-conjugate(P)))
(%i6) A:0+0*%i
(%i7) B:c+0*%i
(%i8) C:u+%i*v
(%i9) F:u+0*%i
(%i10) E:factor(fHa(B,C,A))
(%o10) -(c*u)/(%i*v-u)
(%i11) H:factor(fz(C,F,B,E))
(%o11) -(%i*u*(%i*v+u-c))/v
(%i12) M:(B+F)/2
(%o12) (u+c)/2
(%i13) N:(C+E)/2
(%o13) ((-(c*u)/(%i*v-u))+%i*v+u)/2
(%i14) L:factor((M+N)/2)
(%o14) -(v^2-%i*u*v-%i*c*v+2*u^2+2*c*u)/(4*(%i*v-u))
(%i15) K:factor(fz(M,N,E,F))
(%o15) (%i*u*v^3+%i*c*v^3-u^2*v^2-2*c*u*v^2+c^2*v^2+%i*u^3*v-%i*c^2*u*v-u^4
-c*u^3+c^2*u^2+c^3*u)
/(2*(%i*v-u)*(v^2+u^2-c^2))
(%i16) X:factor((A-L)/(H-K))
(%o16) (v*(v^2+u^2-c^2)*(v^2-%i*u*v-%i*c*v+2*u^2+2*c*u))
/(2*(u-c)*(v^2+u^2-c*u)*(%i*v^2+u*v+c*v+2*%i*u^2+2*%i*c*u))
(%i17) factor(realpart(X))
(%o17) 0
(%i18) factor(imagpart(X))
(%o18) -(v*(v^2+u^2-c^2))/(2*(u-c)*(v^2+u^2-c*u))
