$EF$ is always parallel to a particular line, so there is only one pair of E and F such that $BE+CF=BC$. Let I be the incenter of ABC, now we prove that $(I, ID)$ intersect AB,AC at 4 points, and out of the 4 points there are 2 points that $BCFE$ is concyclic and $BE+CF=BC$. Let $X,Y$ be the symmetric image of D through the internal angle bisector of B and C. Then X,Y is on $(I, ID)$. and using the ratio $\frac{AX}{AB}=\frac{AY}{AC}$, we get that $XY//BC$. Now let X', Y' be the symmetric image of X,Y through the internal angle bisector of A. Then $X'Y'//EF$ (a familiar result), and X',Y' also on $(I, ID)$. $AX'+AY'=AX+AY=AB+AC-BC \Rightarrow CX'+BY'=BC$. So $X',Y'$ are the $E,F$ we find. Out of any other 4 pairs of points of intersection, it is easily seen that no other pairs than $X',Y'$ satisfy $BCY'X'$ concyclic. Thus the conclusion follows.

Pick a point $J$ on $BC$ such that $BJ = BE$ and $CJ = CF$, or $BE + CF = BC$. Let $I$ be the incenter of triangle $ABC$. It is easily seen that the perpendicular bisectors of $EJ$ and $FJ$ meet at $I$. Now let the circle with center $I$ and radius $IE = IF = IJ$ meet $AB$ and $AC$ again at $L$ and $K$, respectively with $AL > AE$ and $CK > CF$ and meet $BC$ at $J$ and $D'$. We will prove that $D'$ is $D$ in our problem. It is seen that $\angle BD'I = \angle BLI$ (symmetry across $BI$) and $\angle BLI = \angle AEI$, or $\angle BD'I = \angle AEI$, and $BD'IE$ is cyclic. Furthermore, we are also given that $BCFE$ is cyclic. Therefore, $\angle D'IE = \angle CFE$. Note that the center angle $ \angle EIK$ subtending arc $EK$ is twice the angle $\angle EFK = \angle AFE$, or $\angle AIE$ = $ \frac{1}{2} \angle EIK = \angle AFE$. However, $\angle CFE + \angle AFE = 180^\circ$; therefore, $\angle D'IE + \angle AIE = 180^\circ $, and point $D'$ is point $D$ in our problem, or the incenter of triangle $ABC$ is also the circumcenter of triangle $DEF$.

For the only if part draw a line through $D$ parallel to $EF$ which will intersect $AC$ and $AB$ at $K$ and $L$ respectively It is easily observed that $KL =CD$ ($\Delta ADK \equiv \Delta ADB$) $\Rightarrow EL+FK=KL$ .But $\frac{EL}{FK}=\frac{AL}{AK}=\frac{DL}{DK} \Rightarrow EL=DL$ and $FK=DK \Rightarrow \angle FED= 90-C/2$ .Now the circle $DEF$ intersects $AB ,BC , CA$ at $P,Q ,R$ again .We easily get $PE=QD=FR$ .The rest is easy

My solution: Let $ I $ be the incenter of $ \triangle ABC $ . Let $ D', E', F' $ be the projection of $ I $ on $ BC, AB, CA $, respectively . Since $ B, C, E, F $ are concyclic , so from $ AB>AC $ we get $ AF>AE $ , hence $ I $ is the circumcenter of $ \triangle DEF \Longleftrightarrow ID=IE=IF $ $ \Longleftrightarrow Rt\triangle IDD' \cong Rt\triangle IEE' \cong Rt\triangle IFF' \Longleftrightarrow DD'=EE'=FF' $ $ \Longleftrightarrow CF+BE=CF'+BE'=CD'+BD'=BC $ . Q.E.D

Let $I$ be the incenter of triangle $ABC$. If $I$ is the circumcircle of $DEF$. This follows $ID=IE$ and $\angle IBE= \angle IBD$. We have $\frac{IE}{\sin \angle IBE}= \frac{ID}{ \sin \angle IBD}$ so $\frac{IB}{\sin \angle BEI}= \frac{IB}{\sin \angle BDI}$. This follows $\angle BEI= \angle BDI$. Therefore $\triangle BEI= \triangle BDI \; ( \text{A.S.A})$ or $BE=BD$. Similarly, we obtain $FC=CD$. Thus, $BE+CF=BC$. If $BE+CF=BC$. Let $D'$ be a point on $BC$ such that $BD'=BE,D'C=CF$. It is easy to prove that $\triangle IBE= \triangle IBD' \; ( \text{S.A.S})$ so $IE=ID$. Similarly, we get $ID=IF$. Hence, $I$ is the circumcircle of triangle $DFE$.

shinichiman: $sin a= sin b <-> a=b $ is not equivanlent

It is clear that as we move the circle passing through $B,C$, the value $BE+CF$ is monotonic, so there exists exactly one circle passing through $B, C$ such that $BE+CF=BC$. Let $I$ be the incenter. Let $E', F'$ distinct from $B, C$ be defined as $E'=AB\cap (IDB)$ and $F'=AC\cap (IDC)$. By Radical Lemma on $(CF'ID), (BE'ID)$, and the concurrency $A=CF'\cap ID\cap BE'$, $B, C, E, F$ is cyclic so $E'=E$ and $F'=F$. If we let $D'$ be the reflection of $D$ about the perpendicular from $I$ to $BC$, then $BE=BD'$ and $CF=CD'$, so $BE+CF=BC$. Finally, since $\angle IDB=\angle IEA$ and $I$ is the incenter, then $ID=IE$. Similarly, $ID=IF$, so $I$ is the circumcenter of $\triangle DEF$. The conditions are all satisfied so we are almost done. It remains to prove that for all other possible circles through $B, C$, the circumcenter of $\triangle DEF$ is not $I$. But this is easy; if $E, F$ are made closer to $A$ than in the above conditions, then $IE > ID$ contradiction, while if $E, F$ are made further to $A$ than in the above conditions, then $IF > ID$ again contradiction, so we are done.

Note that if $E$ and $F$ are points on $AB$ and $AC$, respectively. $AEIF$ is cyclic if and only if $BE+CF=BC$. This pretty much finishes the problem