IMO Shortlist

Let $\omega_{A}$, $\omega_{B}$ and $\omega_{C}$ meet the bisectors $AI$, $BI$ and $CI$ at $A \neq A'$, $B \neq B'$ and $C \neq C'$ respectively. The key observation is that $A'$, $B'$ and $C'$ do not depend on the particular choice of $D$, $E$ and $F$, provided that $BD + BF = CA$, $CD + CE = AB$ and $AE + AF = BC$ hold true (the last equality follows from the other two). For a proof we need the following fact.

Lemma. Given is an angle with vertex $A$ and measure $\alpha$. A circle $\omega$ through $A$ intersects the angle bisector at $L$ and sides of the angle at $X$ and $Y$. Then $AX + AY = 2AL \cos \frac{\alpha}{2}$.

Proof. Note that $L$ is the midpoint of $\operatorname{arc} \widehat{XLY}$ in $\omega$ and set $XL = YL = u$, $XY = v$. By Ptolemy's theorem $AX \cdot YL + AY \cdot XL = AL \cdot XY$, which rewrites as $(AX + AY)u = AL \cdot v$. Since $\angle LXY = \frac{\alpha}{2}$ and $\angle XLY = 180^\circ - \alpha$, we have $v = 2 \cos \frac{\alpha}{2} u$ by the law of sines, and the claim follows. $\square$

Apply the lemma to $\angle BAC = \alpha$ and the circle $\omega = \omega_{A}$, which intersects $AI$ at $A'$. This gives $2AA' \cos \frac{\alpha}{2} = AE + AF = BC$; by symmetry analogous relations hold for $BB'$ and $CC'$. It follows that $A'$, $B'$ and $C'$ are independent of the choice of $D$, $E$ and $F$, as stated.

We use the lemma two more times with $\angle BAC = \alpha$. Let $\omega$ be the circle with diameter $AI$. Then $X$ and $Y$ are the tangency points of the incircle of $ABC$ with $AB$ and $AC$, and hence $AX = AY = \frac{1}{2}(AB + AC - BC)$. So the lemma yields $2AI \cos \frac{\alpha}{2} = AB + AC - BC$. Next, if $\omega$ is the circumcircle of $ABC$ and $AI$ intersects $\omega$ at $M \neq A$ then $\{X, Y\} = \{B, C\}$, and so $2AM \cos \frac{\alpha}{2} = AB + AC$ by the lemma. To summarize,

$$\begin{equation*} 2AA' \cos \frac{\alpha}{2} = BC, \quad 2AI \cos \frac{\alpha}{2} = AB + AC - BC, \quad 2AM \cos \frac{\alpha}{2} = AB + AC. \tag{*} \end{equation*}$$

These equalities imply $AA' + AI = AM$, hence the segments $AM$ and $IA'$ have a common midpoint. It follows that $I$ and $A'$ are equidistant from the circumcenter $O$. By symmetry $OI = OA' = OB' = OC'$, so $I$, $A'$, $B'$, $C'$ are on a circle centered at $O$.

To prove $OP = OI$, now it suffices to show that $I$, $A'$, $B'$, $C'$ and $P$ are concyclic. Clearly one can assume $P \neq I, A', B', C'$.

We use oriented angles to avoid heavy case distinction. The oriented angle between the lines $l$ and $m$ is denoted by $\angle(l, m)$. We have $\angle(l, m) = -\angle(m, l)$ and $\angle(l, m) + \angle(m, n) = \angle(l, n)$ for arbitrary lines $l$, $m$ and $n$. Four distinct non-collinear points $U$, $V$, $X$, $Y$ are concyclic if and only if $\angle(UX, VX) = \angle(UY, VY)$.

Suppose for the moment that $A'$, $B'$, $P$, $I$ are distinct and noncollinear; then it is enough to check the equality $\angle(A'P, B'P) = \angle(A'I, B'I)$. Because $A$, $F$, $P$, $A'$ are on the circle $\omega_{A}$, we have $\angle(A'P, FP) = \angle(A'A, FA) = \angle(A'I, AB)$. Likewise $\angle(B'P, FP) = \angle(B'I, AB)$. Therefore

$$\angle(A'P, B'P) = \angle(A'P, FP) + \angle(FP, B'P) = \angle(A'I, AB) - \angle(B'I, AB) = \angle(A'I, B'I).$$

By Miquel's theorem the circles $(AEF) = \omega_{A}$, $(BFD) = \omega_{B}$ and $(CDE) = \omega_{C}$ have a common point, for arbitrary points $D$, $E$ and $F$ on $BC$, $CA$ and $AB$. So $\omega_{A}$ passes through the common point $P \neq D$ of $\omega_{B}$ and $\omega_{C}$.

Here we assumed that $P \neq F$. If $P = F$ then $P \neq D, E$ and the conclusion follows similarly (use $\angle(A'F, B'F) = \angle(A'F, EF) + \angle(EF, DF) + \angle(DF, B'F)$ and inscribed angles in $\omega_{A}$, $\omega_{B}$, $\omega_{C}$).

There is no loss of generality in assuming $A'$, $B'$, $P$, $I$ distinct and noncollinear. If $ABC$ is an equilateral triangle then the equalities $(*)$ imply that $A'$, $B'$, $C'$, $I$, $O$ and $P$ coincide, so $OP = OI$. Otherwise at most one of $A'$, $B'$, $C'$ coincides with $I$. If say $C' = I$ then $OI \perp CI$ by the previous reasoning. It follows that $A'$, $B' \neq I$ and hence $A' \neq B'$. Finally $A'$, $B'$ and $I$ are noncollinear because $I$, $A'$, $B'$, $C'$ are concyclic.