Problem: Let $ABC$ be a triangle with incenter $I$.
The circle through $B$ tangent to $AI$ at $I$
meets side $AB$ again at $P$.
The circle through $C$ tangent to $AI$ at $I$
meets side $AC$ again at $Q$.
Prove that $PQ$ is tangent to the incircle of $ABC$.

Solution: Let $E$ and $F$ be the tangency points of the incircle on $AC$ and $AB$.
By angle chasing,
\[ \angle PIF = \angle AIF - \angle AIP
  = \left( 90\dg - \half \angle A \right) - \half \angle B = \half \angle C. \]
Similarly, $\angle EIQ = \half \angle B$.

\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair I = incenter(A, B, C);
pair D = foot(I, B, C);
pair E = foot(I, C, A);
pair F = foot(I, A, B);
pair _P = abs(A-I)*abs(A-I)/abs(A-B) * dir(B-A) + A;
pair _Q = abs(A-I)*abs(A-I)/abs(A-C) * dir(C-A) + A;
pair P = _P;
pair Q = _Q;
filldraw(incircle(A, B, C), opacity(0.1)+blue, blue);
filldraw(A--B--C--cycle, opacity(0.1)+blue, blue);
draw(P--Q, red);
draw(P--I--F, deepgreen);
draw(E--I--Q, deepgreen);
pair T = foot(I, P, Q);
draw(I--T, red+dashed);
draw(A--I, lightblue);
draw(B--I--C, lightblue);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$I$", I, dir(270));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$T$", T, dir(T));

/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
A = dir 110
B = dir 210
C = dir 330
I 270 = incenter A B C
D = foot I B C
E = foot I C A
F = foot I A B
!pair _P = abs(A-I)*abs(A-I)/abs(A-B) * dir(B-A) + A;
!pair _Q = abs(A-I)*abs(A-I)/abs(A-C) * dir(C-A) + A;
P = _P
Q = _Q
incircle A B C / 0.1 blue / blue
A--B--C--cycle / 0.1 blue / blue
P--Q / red
P--I--F / deepgreen
E--I--Q / deepgreen
T = foot I P Q
I--T / red dashed
A--I / lightblue
B--I--C / lightblue
*/
\end{asy}
\end{center}

Let $T_p$, $T_q$ denote the second tangency of $P$, $Q$ to the incircle.
Then $\angle E I T_q = \angle B$ and $\angle T_p I F = \angle C$.
Since $\angle E I F = \angle B + \angle C$, it follows $T_p = T_q$.

\begin{remark*}
  Notice we really only need $\angle PIQ = 90\dg - \half \angle A$
  by essentially the same argument.
\end{remark*}