Let $ABC$ be a triangle with altitudes $AD$, $BE$, $CF$. Prove that $EF$ is parallel to the tangent to the circumcircle of $ABC$ at $A$. (And similarly $DF$ and $DE$ are parallel to the tangents through $B$ and $C$, respectively.)
I'm trying to find a simple direct proof for this, but I can't seem to come up with one.
We shall use directed angles modulo $180^{\circ}$. (See Evan Chen, How to Use Directed Angles, 31 May 2015 or, if you read German, my Orientierte Winkel modulo 180° und eine Lösung der WURZEL -Aufgabe Kappa 22 von Wilfried Haag, for an introduction to this type of angles.)
Let $x$ be the tangent to the circumcircle of $\triangle ABC$ at $A$. Thus, $\measuredangle \left(AC, x\right) = \measuredangle CBA$ (by the well-known theorem stating that a tangent-chord angle equals the corresponding chordal angle).
Let $H$ be the orthocenter of $\triangle ABC$. The points $E$ and $F$ lie on the circle with diameter $AH$ (since both angles $\measuredangle AEH$ and $\measuredangle AFH$ are $90^{\circ}$). Thus, the points $A$, $E$, $F$ and $H$ lie on one circle. Hence, the chordal angle theorem yields \begin{align} \measuredangle AEF & = \measuredangle AHF = \measuredangle \left(AH, HC\right) = \underbrace{\measuredangle \left(AH, BC\right)}_{=90^\circ \text{ (since $AH \perp BC$)}} + \underbrace{\measuredangle \left(BC, AB\right)}_{=\measuredangle CBA} + \underbrace{\measuredangle \left(AB, HC\right)}_{=90^\circ \text{ (since $HC \perp AB$)}} \\ &= 90^\circ + \measuredangle CBA + 90^\circ = \underbrace{180^\circ}_{=0^\circ} + \measuredangle CBA = \measuredangle CBA . \end{align} Hence, $\measuredangle \left(AC, EF\right) = \measuredangle AEF = \measuredangle CBA = \measuredangle \left(AC, x\right)$, so that $EF \parallel x$, qed. $\blacksquare$
In the language of triangle geometry, lines parallel to $x$ and $EF$ are said to be antiparallel to $BC$ with respect to $\triangle ABC$.