Problem: Let $\triangle ABC$ be an acute-angled triangle,
and let $D$ be the foot of the altitude from $C$.
The angle bisector of $\angle ABC$ intersects $CD$ at $E$ and
meets the circumcircle $\omega$ of $\triangle ADE$ again at $F$.
If $\angle ADF = 45^{\circ}$, show that $CF$ is tangent to $\omega$.

Solution: \begin{center}
\begin{asy}
pair A = dir(0);
pair B = dir(180);
pair F = dir(65);
pair D = extension(A, B, F, dir(205));
pair K = F*F;
pair C = extension(D, D+dir(90), B, K);
pair E = extension(B, F, C, D);

filldraw(unitcircle, palecyan+opacity(0.2), heavyblue);
filldraw(circumcircle(A, D, E), palered+opacity(0.2), red+dotted);

filldraw(CP(E,D), pink+opacity(0.2), magenta);
draw(arc(F,abs(E-F),170,310), orange+dashed);

draw(K--A--B--C--D, deepcyan);
draw(C--A, deepcyan+dashed);
draw(B--F--A, deepgreen);
draw(F--D, deepgreen);
draw(C--F, mediumred);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$F$", F, dir(F));
dot("$D$", D, dir(225));
dot("$K$", K, dir(K));
dot("$C$", C, dir(C));
dot("$E$", E, dir(165));
\end{asy}
\end{center}

Let $BC$ meet the circle with diameter $\ol{AB}$ at $K$.
By the conditions of the problem, we have $FK = FE = FA$.
Thus $E$ is the incenter of $\triangle KBA$.

By angle chasing, we can now show that
\[ \angle KFE = 90^{\circ} - \frac12 \angle KEF = \angle BCD, \]
so $KCFE$ is cyclic and thus
\[ \angle CKE = 135^{\circ} \implies \angle CFE = 45^{\circ} \]
as needed.
\begin{remark*}
  One can also realize $CKEF$ is cyclic by noting
  \[ BE \cdot BF = BD \cdot BA = BK \cdot BC. \]
\end{remark*}

\begin{remark*}
  Another approach is to use barycentric coordinates on $\triangle ABK$.
  Letting $a=BK$, $b=AK$, $c=AB$ we have $E = (a:b:c)$, $D = (s-b:s-a:0)$, and
  \[ F = (a(a+c) : -b^2 : c(a+c))
    = (a(a+c) : a^2-c^2 : c(a+c))
  = (a : a-c : c). \]
\end{remark*}