Midpoints and perpendiculars in orthocenter configuration

Question

Let $ABC$ be a triangle, and let $E,F$ be the feet of the altitudes from $B,C$ respectively. Let $L$ be the midpoint of $\overline{EF}$, $M$ the midpoint of $\overline{BF}$, $N$ the midpoint of $\overline{CE}$. Prove that $AL$ is perpendicular to $MN$.

I have a solution by complex bashing, but I'm looking for a synthetic solution. One way would be to show that $AW$ is the radical axis of the circles on diameters $\overline{BF}$ and $\overline{CE}$, since it is easy to see that $A$ has equal powers with respect to these circles. However, I haven't really made progress from here.

Answer 1

$\overline{ML}$ is a midpoint segment of $\triangle BFE$, to that $$\overleftrightarrow{ML}\parallel\overleftrightarrow{BE} \quad\to\quad \overleftrightarrow{ML}\perp\overleftrightarrow{AC}$$ Likewise, $$\overleftrightarrow{NL}\perp\overleftrightarrow{AB}$$ Therefore, $L$, as the intersection of two altitudes of $\triangle AMN$, is its orthocenter. Necessarily, $\overleftrightarrow{AL}$ is the third altitude, giving the result. $\square$

Answer 2

We may exploit the definition of a perpendicular as the locus of points for which a difference of squared distances is constant. Let $W=AL\cap MN$ and let $P$ be the midpoint of $BC$. $PNLM$ is so the Varignon parallelogram of the cyclic quadrilateral $CEFB$. By Thales' theorem and the Pythagorean theorem: $$LM^2-LN^2 = \frac{1}{4}\left(BE^2-CF^2\right) = \frac{1}{4}\left(BF^2-CE^2\right)=BM^2-CN^2 $$ Now it is straightforward to check that $AM^2-AN^2$ equals the same thing, hence $AL\perp MN$ as wanted. Even easier: it is enough to consider that $AMPN$ is a cyclic quadrilateral and $AP$ is a diameter of its circumcircle.

Answer 3

Let $\omega_1$, $\omega_2$, and $\omega$ be the circles with diameters $BF$, $EC$, and $\omega$, respectively.

Since $B,C,E,F \in \omega$, we have, by the power of point, $$AF \cdot AB = AE \cdot AC,$$ therefore $A$ lies on the radical axis of $\omega_1, \omega_2$.

Let $X$ and $Y$ be the orthogonal projections of $B$ and $C$ onto $EF$, respectively. By definition, $X \in \omega_1$ and $Y \in \omega_2$. Denote the midpoint of $BC$ by $P$. Note that $PF=PE$, so $PL \perp EF$. We have therefore $BX \parallel PL \parallel CY$. Since $P$ is the midpoint of $BC$, $L$ is the midpoint of $XY$. Therefore $$LF \cdot LX = LE \cdot LY.$$ Thus $L$ lies on the radical axis of $\omega_1, \omega_2$.

We have proved that $AL$ is radical axis of $\omega_1, \omega_2$, and therefore $AL$ is perpendicular to the line passing through the centers of this circles, i.e. to the line $MN$.

Answer 4

Observe that $NL$ is the midsegment in triangle $EFC$ parallel to $CF$ and half of it in length. Therefore $NL$ is perpendicular to $AB$. Analogously, $ML$ is the midsegment in triangle $EFB$ parallel to $BE$ and half of it in length. Therefore $ML$ is perpendicular to $CA$. Hence the lines $NL$ and $ML$ are the altitudes of triangle $AMN$ through vertices $N$ and $M$ respectively. Therefore point $L$ is the orthocenter of triangle $AMN$ and hence $AL$ is the third altitude which means that $AL$ is perpendicular to $MN$.