Perpendiculars and midpoints in orthocenter configuration

Question

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 $\overline{BF}, \overline{CE}$ be $M,N$. Let $MN$ meet $EF$ at $K$, and let $L$ be the midpoint of $\overline{MN}$. Prove that $AL\perp HK$.

Although there are lots of perpendiculars, Cartesian coordinates or vectors just don't seem nice enough. I can't seem to pull of a bash with barycentric coordinates either. I'd really welcome a synthetic proof.

Here's some progress on the bary bash:

We have $E=\left(\frac{a^2+b^2-c^2}{2b^2},0,\frac{-a^2+b^2+c^2}{2c^2}\right)$ and $F=\left(\frac{a^2-b^2+c^2}{2c^2},\frac{-a^2+b^2+c^2}{2c^2},0\right)$. From this, we get $M=\left(a^2-b^2+c^2:-a^2+b^2+3c^2:0\right)$ and $N=\left(a^2+b^2-c^2:0:-a^2+3b^2+c^2\right)$. The equation of line $EF$ is $$\left(-a^2+b^2+c^2\right)x-\left(a^2-b^2+c^2\right)y-\left(a^2+b^2-c^2\right)z=0,$$ and the equation of line $MN$ is $$\left(-a^2+b^2+3c^2\right)\left(-a^2+3b^2+c^2\right)x-\left(a^2-b^2+c^2\right)\left(-a^2+3b^2+c^2\right)y-\left(a^2+b^2-c^2\right)\left(-a^2+b^2+3c^2\right)z=0.$$ Intersecting these lines to get $K$ seems horrid.

If I had the coordinates of $K$, then verifying that $AL\perp HK$ isn't too bad.

EDIT: I've found the coordinates of $K$. Using Conway notation, we have $$K=\left(\frac{S_{BC}}{2\left(S^2+S_A^2\right)},\frac{S_{BC}+S_{CA}}{2\left(S^2-2S_{AB}\right)-2S_B^2},\frac{S_{BC}+S_{AB}}{2\left(S^2-2S_{CA}\right)-2S_C^2}\right).$$

However, applying EFFT to prove the perpendicularity is also proving horrific.

Answer 1

$\def\conj#1{\overline{#1}}\def\i{\mathbf{i}}$

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))