Problem: Let $ABC$ be a triangle with $\angle B > 90\dg$
and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$.
Can line $EF$ be tangent to the $A$-excircle?

Solution: We show it is not possible, by contradiction
(assuming $EF$ is indeed tangent).
Thus $BECF$ is a convex cyclic quadrilateral
inscribed in a circle with diameter $\ol{BC}$.
Note also that the $A$-excircle lies on the opposite side
from $A$ as line $EF$, since $A$, $E$, $C$ are collinear
in that order.

\paragraph{First solution by similarity.}
Note that $\triangle AEF$ is similar to $\triangle ABC$
(and oppositely oriented).
However, since they have the same $A$-exradius,
it follows they are congruent.

\begin{center}
\begin{asy}
size(6cm);
pair B = dir(213);
pair C = dir(47);
pair E = dir(115);
pair F = dir(280);

pair V = extension(F, E, B, C);
pair W = incenter(F, V, C);
pair J = 3.6*W-2.6*V;

pair T = foot(J, B, C);
pair A = extension(B, F, E, C);

filldraw(CP(J, T), opacity(0.1)+deepgreen, deepgreen);

pair U = OP(CP(J, T), CP(midpoint(B--J), J));
pair V = IP(CP(J, T), CP(midpoint(E--J), J));
fill(A--B--C--cycle, rgb(0.9,1,1));

draw(B--E, lightblue);
draw(C--F, lightblue);
draw(U--B--A--V, blue);
draw(B--C, blue);
draw(F--E, red);
draw(unitcircle, lightblue+dotted);

dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$A$", A, dir(A));

/* TSQ Source:

!size(6cm);
B = dir 213
C = dir 47
E = dir 115
F = dir 280

V := extension F E B C
W := incenter F V C
J := 3.6*W-2.6*V

T := foot J B C
A = extension B F E C

CP J T 0.1 deepgreen / deepgreen

U := OP CP J T CP midpoint B--J J
V := IP CP J T CP midpoint E--J J
!fill(A--B--C--cycle, rgb(0.9,1,1));

B--E lightblue
C--F lightblue
U--B--A--V blue
B--C blue
F--E red
unitcircle lightblue dotted

*/
\end{asy}
\end{center}

Consequently we get $EF = BC$.
But this implies $BFCE$ is a rectangle, contradiction.

\paragraph{Second length solution by tangent lengths.}
By $t(\bullet)$ we mean the length of the tangent from $P$
to the $A$-excircle.
It is a classical fact for example that $t(A) = s$.
The main idea is to use the fact that
\[ a \cos A = EF = t(E) + t(F). \]
Here $EF = a \cos A$ follows from the extended law of sines
applied to the circle with diameter $\ol{BC}$,
since there we have $EF = BC \sin \angle ECF = a \sin \angle ACF = a \cos A$.
We may now compute
\begin{align*}
  t(E) &= t(A) - AE = s - c \cos A \\
  t(F) &= t(A) - AF = s - b \cos A.
\end{align*}
Therefore,
\begin{align*}
  a \cos A = 2s - (b+c) \cos A \implies (a+b+c) \cos A &= 2s \\
  \implies \cos A &= 1.
\end{align*}
This is an obvious contradiction.

\begin{remark*}
  On the other hand, there really is an \emph{equality case}
  with $A$ being some point at infinity (meaning $\cos A = 1$).
  So, this problem is ``sharper'' than one might expect;
  the answer is not ``obviously no''.
\end{remark*}

\paragraph{Third solution by Pitot and trigonometry.}
In fact, the $t(\bullet)$ notation from the previous
solution gives us a classical theorem
once we note the $A$-excircle is tangent to all four lines
$EF$, $BC$, $BF$ and $CE$:
\begin{claim*}
  [Pitot theorem]
  We have $BF + EF = BC + CE$.
\end{claim*}
\begin{proof}
  Here is a proof for completeness.
  By $t(B)$ we mean the length of the tangent from $B$ to
  the $A$-excircle, and define $t(C)$, $t(E)$, $t(F)$ similarly.
  Then
  \begin{align*}
    BF &= t(B) - t(F) & EF &= t(E) + t(F) \\
    BC &= t(B) + t(C) & CE &= t(E) - t(C)
  \end{align*}
  and summing gives the result.
\end{proof}
\begin{center}
\begin{asy}
pair B = dir(180);
pair C = -B;
pair E = dir(40);
pair F = -E;

pair X = B+2*dir(F-B);
pair Y = C+2*dir(C-E);
pair J = extension(C, X, B, Y);

filldraw(CP(J, foot(J, B, C)), opacity(0.1)+deepgreen, deepgreen);
pair K1 = B+dir(E-C)*1.7;
pair K2 = C+dir(E-C)*1.7;

pair U = foot(J, B, F);
pair V = foot(J, C, E);
label("$A$", midpoint(K1--K2), dir(C-E));

filldraw(K1--B--C--K2--cycle, opacity(0.1)+lightcyan, invisible);
draw(K1--B--C--K2, blue);
draw(F--E, red);
draw(B--U, blue);
draw(C--V, blue);
draw(B--E, blue);
draw(C--F, blue);

dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$J$", J, dir(-90));

/* TSQ Source:

B = dir 180
C = -B
E = dir 40
F = -E

X := B+2*dir(F-B)
Y := C+2*dir(C-E)
J = extension C X B Y R-90

CP J foot J B C 0.1 deepgreen / deepgreen
K1 := B+dir(E-C)*1.7
K2 := C+dir(E-C)*1.7

U := foot J B F
V := foot J C E
!label("$A$", midpoint(K1--K2), dir(C-E));

K1--B--C--K2--cycle 0.1 lightcyan / invisible
K1--B--C--K2 blue
F--E red
B--U blue
C--V blue
B--E blue
C--F blue

*/
\end{asy}
\end{center}
We now calculate all the lengths using trigonometry:
\begin{align*}
  BC &= a \\
  BF &= a \cos(180\dg-B) = a \cos (A+C) \\
  CE &= a \cos C \\
  EF &= BC \sin \angle ECF = a \sin \angle ACF = a \cos A.
\end{align*}
Thus, we apparently have
\[ \cos(A+C) + \cos A = 1 + \cos C \]
but this is impossible since $\cos(A+C) < \cos C$
(since $A+C = 180-B < 90\dg$) and $\cos A < 1$.

\paragraph{Fourth solution by Pitot and Ptolemy (Evan Chen).}
We give a trig-free way to finish from Pitot's theorem
\[ BF+EF = BC+CE. \]
Assume that $x = BF$, $y = CE$, and $BC = 1$;
then the above relation becomes
\[ 1 + y - x = BC + CE - BF = EF = EF \cdot 1 = xy + \sqrt{(1-x^2)(1-y^2)} \]
with the last step by Ptolemy's theorem.
This rearranges to give
\[ (1+y)(1-x) = \sqrt{(1-x^2)(1-y^2)}
  \implies \frac{1+y}{1-y} = \frac{1+x}{1-x}
  \implies x = y \]
but that means $BECF$ is a rectangle:
contradicting the fact that lines $BE$ and $CF$ meet at a point $A$.

\paragraph{Fifth solution, by angle chasing only!}
Let $J$ denote the $A$-excenter.
Then $J$ should be the intersection
of the internal bisectors of $\angle FEC$ and $\angle FBC$,
so it is the midpoint of arc $\widehat{FC}$
on the circle with diameter $\ol{BC}$.

\begin{center}
\begin{asy}
pair B = dir(180);
pair C = -B;
pair E = dir(40);
pair F = -E;

pair X = B+2*dir(F-B);
pair Y = C+2*dir(C-E);
pair J = extension(C, X, B, Y);

filldraw(CP(J, foot(J, B, C)), opacity(0.1)+deepgreen, deepgreen);
pair K1 = B+dir(E-C)*1.7;
pair K2 = C+dir(E-C)*1.7;

pair U = foot(J, B, F);
pair V = foot(J, C, E);
label("$A$", midpoint(K1--K2), dir(C-E));

filldraw(K1--B--C--K2--cycle, opacity(0.1)+lightcyan, invisible);
draw(K1--B--C--K2, blue);
draw(F--E, red);
draw(B--U, blue);
draw(C--V, blue);
draw(B--E, blue);
draw(C--F, blue);

filldraw(circumcircle(B, C, J), opacity(0.1)+yellow, orange+dashed);
draw(B--J--E, orange);

dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$J$", J, dir(-90));

/* TSQ Source:

B = dir 180
C = -B
E = dir 40
F = -E

X := B+2*dir(F-B)
Y := C+2*dir(C-E)
J = extension C X B Y R-90

CP J foot J B C 0.1 deepgreen / deepgreen
K1 := B+dir(E-C)*1.7
K2 := C+dir(E-C)*1.7

U := foot J B F
V := foot J C E
!label("$A$", midpoint(K1--K2), dir(C-E));

K1--B--C--K2--cycle 0.1 lightcyan / invisible
K1--B--C--K2 blue
F--E red
B--U blue
C--V blue
B--E blue
C--F blue

circumcircle B C J 0.1 yellow / orange dashed
B--J--E orange

*/
\end{asy}
\end{center}

But now we get $\angle BJC = 90\dg$ from $J$ lying on this circle.
Yet $\angle BJC = 90\dg - \half \angle A$ in general,
so $\angle A = 0\dg$ which is impossible.

\paragraph{Sixth solution (Zuming Feng).}
This is similar to the preceding solution,
but phrased using contradiction and inequalities.
We let $X$ and $Y$ denote the tangency points of the $A$-excircle
on lines $AB$ and $AC$.
Moreover, let $J$ denote the $A$-excenter.

\begin{center}
\begin{asy}
size(8cm);
pair B = dir(213);
pair C = dir(47);
pair E = dir(115);
pair F = dir(280);

pair V = extension(F, E, B, C);
pair W = incenter(F, V, C);
pair J = 3.6*W-2.6*V;

pair T = foot(J, B, C);
pair A = extension(B, F, E, C);

filldraw(CP(J, T), opacity(0.1)+deepgreen, deepgreen);

pair X = OP(CP(J, T), CP(midpoint(B--J), J));
pair Y = IP(CP(J, T), CP(midpoint(E--J), J));
fill(A--B--C--cycle, rgb(0.9,1,1));

draw(B--E, lightblue);
draw(C--F, lightblue);
draw(X--B--A--Y, blue);
draw(B--C, blue);
draw(F--E, red);
draw(unitcircle, lightblue+dotted);

dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$J$", J, dir(J));
dot("$A$", A, dir(A));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));

/* TSQ Source:

!size(6cm);
B = dir 213
C = dir 47
E = dir 115
F = dir 280

V := extension F E B C
W := incenter F V C
J = 3.6*W-2.6*V

T := foot J B C
A = extension B F E C

CP J T 0.1 deepgreen / deepgreen

X = OP CP J T CP midpoint B--J J
Y = IP CP J T CP midpoint E--J J
!fill(A--B--C--cycle, rgb(0.9,1,1));

B--E lightblue
C--F lightblue
X--B--A--Y blue
B--C blue
F--E red
unitcircle lightblue dotted

*/
\end{asy}
\end{center}

Note that $AB > AE$ and $AX = AY$, therefore $BX < EY$.
By considering the right triangles $XBJ$ and $YEJ$
(which both have $JX = JY$),
we conclude $\tan \angle XBJ > \tan \angle YEJ$, thus
\[ \angle XBJ > \angle YEJ. \]
However, if line $EF$ was actually tangent to the $A$-excircle,
we would have
\[ 2\angle XBJ = \angle XBC = \angle FBC = \angle FEC
  = \angle FEY = 2 \angle JEY \]
which is a contradiction.

\paragraph{Seventh solution, by complex numbers, for comedic effect (Evan Chen).}
Let us denote the tangency points of the $A$-excircle
with sides $BC$, $CA$, $AB$ as $x$, $y$, $z$.
Assume moreover that line $EF$ is tangent to the $A$-excircle
at a point $P$.

Also, for brevity let $s = xy+yz+zx$.
Then, we have
\begin{align*}
  E = \frac{2py}{p+y} &= \half(b+y+y-y^2 \ol b)
  = \frac{zx}{z+x} + y - \frac{y^2}{z+x} \\
  \implies \frac{2}{\frac1p+\frac1y} &= \frac{xy+xz+zx-y^2}{z+x}
  \implies \frac{\frac1p+\frac1y}{2} = \frac{x+z}{s-y^2}. \\
  \intertext{Similarly by considering the point $F$,}
  \frac{\frac1p+\frac1z}{2} &= \frac{x+y}{s-z^2}. \\
  \intertext{Thus we can eliminate $P$ and obtain}
  \implies \frac{\frac1y-\frac1z}{2} &= \frac{x+z}{s-y^2} - \frac{x+y}{s-z^2}
  = \frac{-s(y-z) + x(y^2-z^2) + (y^3-z^3)}{(s-y^2)(s-z^2)} \\
  \iff \frac{1}{2yz} &= \frac{s - x(y+z) - (y^2+yz+z^2)}{(s-y^2)(s-z^2)}
    = \frac{-(y^2+z^2)}{(s-y^2)(s-z^2)} \\
  \iff 0 &= (s-y^2)(s-z^2) + 2yz(y^2+z^2) \\
  &= \left[ x(y+z) + y(z-y) \right]
    \left[ x(y+z) + z(y-z) \right] + 2yz(y^2+z^2) \\
  &= x^2(y+z)^2 - (y-z)^2 \cdot x(y+z) + yz(2y^2+2z^2 - (y-z)^2) \\
  &= x^2(y+z)^2 - (y-z)^2 \cdot x(y+z) + yz(y+z)^2 \\
  &= xyz(y+z) \left[ \frac xy + \frac xz
    - \frac yz - \frac zy + 2
    + \frac yx + \frac zx \right].
\end{align*}
However, $\triangle XYZ$ is obtuse with $\angle X > 90\dg$,
we have $y+z \neq 0$.
Note that
\begin{align*}
  \tfrac xy + \tfrac yx &= 2\opname{Re} \tfrac xy = 2 \cos(2 \angle XZY) \\
  \tfrac xz + \tfrac zx &= 2\opname{Re} \tfrac xz = 2 \cos(2 \angle XYZ) \\
  \tfrac yz + \tfrac zy &= 2\opname{Re} \tfrac yz < 2
\end{align*}
and since $\cos(2\angle XZY) + \cos(2\angle XYZ) > 0$
(say by sum-to-product), we are done.