Problem: Let $ABC$ be a triangle such that $\angle CAB > \angle ABC$, and let $I$ be its incenter.
Let $D$ be the point on segment $BC$ such that $\angle CAD = \angle ABC$.
Let $\omega$ be the circle tangent to $AC$ at $A$ and passing through $I$.
Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $ABC$.
Prove that the angle bisectors of $\angle DAB$ and $\angle CXB$
intersect at a point on line $BC$.

Solution: Here is a cross-ratio/trig solution:
rare for me to do one of these,
but this problem called out to me that way.
As usual, let $\alpha = \angle BAC$, etc.

Let $T$ be the foot of the bisector of $\angle BAD$ on $\ol{BD}$,
so that
\[ \frac{TB}{TC} = \frac{AB \sin \angle BAT}{AC \sin \angle CAT}
  = \frac{AB \sin \frac{\alpha-\beta}{2}}{AC \sin \frac{\alpha+\beta}{2}}. \]
Also, call $(ABC)$ by $\Gamma$ and let ray $XI$ meet $\Gamma$ again at $W$,
meaning that $\angle WXA = \angle IXA = \angle IAC = \frac{\beta}{2}$,
since we assume $\ol{AC}$ was tangent to $(IAX)$.
Thus arc $\widehat{AW}$ also has measure $\beta$.

\begin{center}
\begin{asy}
pair A = dir(50);
pair B = dir(210);
pair C = dir(330);
pair I = incenter(A, B, C);
pair M_a = circumcenter(B, I, C);
pair M_b = circumcenter(C, I, A);
pair M_c = circumcenter(A, I, B);

pair W = A*B/M_a;
pair X = -W+2*foot(origin, I, W);
pair D = extension(B, C, A, C*C/A);
filldraw(unitcircle, opacity(0.1)+lightcyan, lightblue);
draw(A--B--C--cycle, lightblue);
draw(X--W, lightred);
draw(B--M_b, lightred);
draw(C--M_c, lightred);
draw(A--M_a, lightred);

filldraw(W--M_b--A--M_c--cycle, opacity(0.1)+yellow, deepgreen);
filldraw(X--B--M_a--C--cycle, opacity(0.1)+yellow, deepgreen);
draw(A--D, lightblue);
pair T = extension(A, incenter(A, B, D), B, D);
draw(X--T--A, dashed+brown);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$I$", I, dir(I));
dot("$M_a$", M_a, dir(M_a));
dot("$M_b$", M_b, dir(M_b));
dot("$M_c$", M_c, dir(M_c));
dot("$W$", W, dir(W));
dot("$X$", X, dir(X));
dot("$D$", D, dir(D));
dot("$T$", T, dir(T));

/* TSQ Source:

A = dir 50
B = dir 210
C = dir 330
I = incenter A B C
M_a = circumcenter B I C
M_b = circumcenter C I A
M_c = circumcenter A I B

W = A*B/M_a
X = -W+2*foot origin I W
D = extension B C A C*C/A
unitcircle 0.1 lightcyan / lightblue
A--B--C--cycle lightblue
X--W lightred
B--M_b lightred
C--M_c lightred
A--M_a lightred

W--M_b--A--M_c--cycle 0.1 yellow / deepgreen
X--B--M_a--C--cycle 0.1 yellow / deepgreen
A--D lightblue
T = extension A incenter A B D B D
X--T--A dashed brown

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

Now,
\begin{align*}
  \frac{XB}{XC} &= -(BC;XM_a)_\Gamma \overset{I}{=} -(M_b M_c; W A)_\Gamma \\
  &= -\frac{\sin \half \mathrm{m}\widehat{M_b W}}{\sin \half \mathrm{m}\widehat{M_c W}}
    \div \frac{\sin \half \mathrm{m}\widehat{M_b A}}{\sin \half \mathrm{m}\widehat{M_c A}}
  = \frac{\sin\frac{\alpha-\beta}{2}}
    {\sin\frac{\alpha +\gamma}{2}}
    \div \frac{\sin \frac{\beta}{2}}{\sin \frac{\gamma}{2}} \\
\end{align*}
Therefore,
\[ \frac{XB}{XC} \div \frac{TB}{TC}
  = \frac{\sin\frac{\alpha+\beta}{2} \sin\frac{\gamma}{2}}%
  {\sin\frac{\alpha+\gamma}{2} \sin\frac{\beta}{2}}
  \div \frac{AB}{AC} \\
  = \frac{2\cos \frac{\gamma}{2} \sin \frac{\gamma}{2}}%
  {2\cos \frac{\beta}{2} \sin \frac{\beta}{2}} \div \frac{AB}{AC} = 1 \]
the last step being the law of sines $AB/AC = \sin\gamma/\sin\beta$.