Problem: Let $ABCDEF$ be a convex hexagon such that $\angle A = \angle C = \angle E$ and $\angle B = \angle D = \angle F$
and the (interior) angle bisectors of $\angle A$, $\angle C$, and $\angle E$ are concurrent.
Prove that the (interior) angle bisectors of $\angle B$, $\angle D$, and $\angle F$ must also be concurrent.

Solution: In general, if hexagon $ABCDEF$ has $\angle A = \angle C = \angle E$ and $\angle B = \angle D = \angle F$,
then its sides can be extended to form two
equilateral triangles $PQR$ and $XYZ$, as shown.

\begin{center}
\begin{asy}
size(5cm);
pair P = dir(90);
pair Q = dir(210);
pair R = dir(330);

pair O = (0.1,0.2);
pair z = dir(-40);

pair Y = O+z*(P-O);
pair X = O+z*(R-O);
pair Z = O+z*(Q-O);

filldraw(P--Q--R--cycle, opacity(0.1)+lightblue, blue);
filldraw(X--Y--Z--cycle, opacity(0.1)+lightcyan, deepcyan);

pair A = extension(Q, P, Y, Z);
pair B = extension(P, R, Y, Z);
pair C = extension(P, R, X, Y);
pair D = extension(R, Q, X, Y);
pair E = extension(R, Q, X, Z);
pair F = extension(Q, P, X, Z);
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$R$", R, dir(R));
dot("$Y$", Y, dir(Y));
dot("$X$", X, dir(X));
dot("$Z$", Z, dir(Z));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
\end{asy}
\end{center}

The problem is solved upon proving the following claim.

\begin{claim*}
  The angle bisectors of $\angle A$, $\angle C$, $\angle E$ are concurrent if and only if
  the unique spiral similarity sending $PQR$ to $YZX$ is a rotation;
  equivalently, the two triangles are congruent.
\end{claim*}

\begin{center}
\begin{asy}
pair P = dir(90);
pair Q = dir(210);
pair R = dir(330);

pair O = (0.1,0.2);
pair z = dir(-40);

pair Y = O+z*(P-O);
pair X = O+z*(R-O);
pair Z = O+z*(Q-O);

filldraw(P--Q--R--cycle, opacity(0.1)+lightblue, blue);
filldraw(X--Y--Z--cycle, opacity(0.1)+lightcyan, deepcyan);

pair A = extension(Q, P, Y, Z);
pair B = extension(P, R, Y, Z);
pair C = extension(P, R, X, Y);
pair D = extension(R, Q, X, Y);
pair E = extension(R, Q, X, Z);
pair F = extension(Q, P, X, Z);

draw(circumcircle(O, A, C), deepgreen);
draw(circumcircle(O, C, E), deepgreen);
draw(circumcircle(O, E, A), deepgreen);
draw(A--O, red);
draw(C--O, red);
draw(E--O, red);

label("$2\theta$", A, 5*dir(40), orange);
label("$2\theta$", C, 5*dir(280), orange);
label("$2\theta$", E, 5*dir(160), orange);

dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$R$", R, dir(R));
dot("$O$", O, dir(O));
dot("$Y$", Y, dir(Y));
dot("$X$", X, dir(X));
dot("$Z$", Z, dir(Z));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));

/* TSQ Source:

P = dir 90
Q = dir 210
R = dir 330

O = (0.1,0.2)
z := dir -40

Y = O+z*(P-O)
X = O+z*(R-O)
Z = O+z*(Q-O)

P--Q--R--cycle 0.1 lightblue / blue
X--Y--Z--cycle 0.1 lightcyan / deepcyan

A = extension Q P Y Z
B = extension P R Y Z
C = extension P R X Y
D = extension R Q X Y
E = extension R Q X Z
F = extension Q P X Z

circumcircle O A C deepgreen
circumcircle O C E deepgreen
circumcircle O E A deepgreen
A--O red
C--O red
E--O red

!label("$2\theta$", A, 5*dir(40), orange);
!label("$2\theta$", C, 5*dir(280), orange);
!label("$2\theta$", E, 5*dir(160), orange);

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

\begin{proof}[Proof of ``if'' direction]
  Let $O$ be the center of the rotation.
  Then $O$ is equidistant form $\ol{YZ}$ and $\ol{PQ}$ by rotation;
  similarly for the other pairs of sides.
  So the bisectors meet at $O$.
\end{proof}

\begin{proof}[Proof of ``only if'' direction]
  Let $O$ be the concurrency point.
  Let $\angle PAB = \angle RCD = \angle QEF = 2\theta$
  Since $\angle APC = \angle AYC = 60\dg$ the quadrilateral $PACY$ is cyclic.
  But $\angle PAO + \angle PCO = (90\dg+\theta) + (90\dg-\theta) = 180\dg$, so $PAOCY$ is cyclic.

  Now $\angle PAO = \angle YCO \implies OP = OY$, and $\angle POY = 2\theta$.
  So $O$ is the center of rotation from $\triangle PQR$ to $\triangle YXZ$ with angle $2\theta$.
\end{proof}