Problem: Let $ABC$ be a triangle with incenter $I$
and excenters $I_a$, $I_b$, $I_c$ opposite $A$, $B$, and $C$, respectively.
Given an arbitrary point $D$ on the circumcircle of $\triangle ABC$
that does not lie on any of the lines $II_a$, $I_bI_c$, or $BC$,
suppose the circumcircles of $\triangle DII_a$ and $\triangle DI_bI_c$
intersect at two distinct points $D$ and $F$.
If $E$ is the intersection of lines $DF$ and $BC$,
prove that $\angle BAD = \angle EAC$.

\end{enumerate}

Solution: Here are two approaches.
\begin{center}
\begin{asy}
size(9cm);

pair A = dir(115);
pair B = dir(210);
pair C = dir(330);
pair D = dir(240);
pair E = extension(B, C, A, B*C/D);
pair I = incenter(A, B, C);
pair M = dir(270);
pair I_a = 2*M-I;
pair I_b = extension(B, I, C, I_a);
pair I_c = extension(C, I, B, I_a);
pair F = -D+2*foot(circumcenter(D, I, I_a), D, E);
draw(D--F, deepgreen);

filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, blue);
draw(circumcircle(D, I, I_a), lightred);
draw(circumcircle(D, I_b, I_c), lightred);

draw(A--I_a, blue);
draw(I_a--I_b--I_c--cycle, gray);
draw(D--A--E, deepcyan+dashed);

dot("$A$", A, dir(A));
dot("$B$", B, dir(170));
dot("$C$", C, dir(350));
dot("$D$", D, dir(D));
dot("$E$", E, dir(280));
dot("$I$", I, dir(160));
dot("$M$", M, dir(225));
dot("$I_a$", I_a, dir(I_a));
dot("$I_b$", I_b, dir(I_b));
dot("$I_c$", I_c, dir(I_c));
dot("$F$", F, dir(F));

/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(9cm);
A = dir 115
B 170 = dir 210
C 350 = dir 330
D = dir 240
E 280 = extension B C A B*C/D
I 160 = incenter A B C
M 225 = dir 270
I_a = 2*M-I
I_b = extension B I C I_a
I_c = extension C I B I_a
F = -D+2*foot (circumcenter D I I_a) D E
D--F deepgreen
unitcircle / 0.1 lightcyan / blue
A--B--C--cycle / 0.1 lightcyan / blue
circumcircle D I I_a / lightred
circumcircle D I_b I_c / lightred
A--I_a / blue
I_a--I_b--I_c--cycle / gray
D--A--E / deepcyan dashed
*/
\end{asy}
\end{center}

\paragraph{Barycentric coordinates (Carl Schildkraut).}
With reference triangle $\triangle ABC$, set $D = (r:s:t)$.
\begin{claim*}
  The equations of $(DII_a)$ and $(DI_bI_c)$ are, respectively,
  \begin{align*}
    0 &= -a^2yz-b^2zx-c^2xy + (x+y+z)
    \cdot \left( bc x - \frac{bcr}{cs-bt} (cy-bz) \right) \\
    0 &= -a^2yz-b^2zx-c^2xy + (x+y+z)
    \cdot \left( -bc x + \frac{bcr}{cs+bt} (cy+bz) \right).
  \end{align*}
\end{claim*}
\begin{proof}
  Since $D \in (ABC)$, we have $a^2st+b^2tr+c^2rs=0$.
  Now each equation can be verified by direct substitution of three points.
\end{proof}

By EGMO Lemma 7.24, the radical axis is then given by
\[ \ol{DF}: bc x - \frac{bcr}{cs-bt} (cy-bz)
    = -bc x + \frac{bcr}{cs+bt} (cy+bz). \]
Now the point
\[ \left( 0 : \frac{b^2}{s} : \frac{c^2}{t} \right)
  = \left( 0 : b^2t : c^2s \right) \]
lies on line $DF$ by inspection, and is obviously on line $BC$,
hence it coincides with $E$.
This lies on the isogonal of $\ol{AD}$ (by EGMO Lemma 7.6), as needed.

\paragraph{Synthetic approach (Anant Mudgal).}
Focus on just $(DII_a)$.
Let $P$ be the second intersection of $(DII_a)$ with $(ABC)$,
and let $M$ be the midpoint of minor arc $\arc{BC}$.
Then by radical axis, lines $AM$, $DP$, and $BC$ are concurrent at a point $K$.

Let $E' = \ol{PM} \cap \ol{BC}$.

\begin{center}
\begin{asy}
size(9cm);

pair A = dir(140);
pair B = dir(210);
pair C = dir(330);
pair D = dir(240);
pair E_prime = extension(B, C, A, B*C/D);
pair I = incenter(A, B, C);
pair M = dir(270);
pair I_a = 2*M-I;
pair K = extension(A, I, B, C);
pair P = extension(D, K, M, E_prime);
pair X = 2*M-E_prime;

filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, blue);
filldraw(circumcircle(D, I, I_a), opacity(0.1)+yellow, blue);
draw(A--I_a, blue);
draw(D--P--X, orange);
filldraw(X--I--E_prime--I_a--cycle, opacity(0.1)+magenta, red);
filldraw(circumcircle(B, I, C), opacity(0.1)+gray, gray);

draw(D--A--E_prime, deepcyan+dashed);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(350));
dot("$D$", D, dir(D));
dot("$E'$", E_prime, dir(325));
dot("$I$", I, dir(160));
dot("$M$", M, dir(M));
dot("$I_a$", I_a, dir(I_a));
dot("$K$", K, dir(325));
dot("$P$", P, dir(P));
dot("$X$", X, dir(X));

/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(9cm);
A = dir 140
B = dir 210
C 350 = dir 330
D = dir 240
E' 325 = extension B C A B*C/D
I 160 = incenter A B C
M = dir 270
I_a = 2*M-I
K 325 = extension A I B C
P = extension D K M E'
X = 2*M-E'
unitcircle / 0.1 lightcyan / blue
A--B--C--cycle / 0.1 lightcyan / blue
circumcircle D I I_a / 0.1 yellow / blue
A--I_a / blue
D--P--X / orange
X--I--E'--I_a--cycle / 0.1 magenta / red
circumcircle B I C / 0.1 gray / gray
D--A--E' / deepcyan dashed
*/
\end{asy}
\end{center}

\begin{claim*}
  We have $\dang BAD = \dang E'AC$.
\end{claim*}
\begin{proof}
  By shooting lemma, $AKE'P$ is cyclic, so
  \[ \dang KAE' = \dang KPE' = \dang DPM = \dang DAM. \qedhere \]
\end{proof}

\begin{claim*}
  The power of point $E'$ with respect to $(DII_a)$ is $2 E'B \cdot E'C$.
\end{claim*}
\begin{proof}
  Construct parallelogram $IE'I_aX$.
  Since $MI^2 = ME' \cdot MP$, we can get
  \[ \dang XI_aI = \dang I_aIE' = \dang MIE' = \dang MPI = \dang XPI. \]
  Hence $X$ lies on $(DII_a)$,
  and $E'X \cdot E'P = 2 E'M \cdot E'P = 2 E'B \cdot E'C$.
\end{proof}

Repeat the argument on $(DI_bI_c)$;
the same point $E'$ (because of the first claim)
then has power $2 E'B \cdot E'C$ with respect to $(DI_bI_c)$.
Hence $E'$ lies on the radical axis of $(DII_a)$ and $(DI_bI_c)$, ergo $E'=E$.
The first claim then solves the problem.