Making O the origin of a complex plane, we can prove that the line EF pass through (A+B+C+D)/2=D/2 (this is a shorlist problem). So the point of intersection of EF and OD is the midpoint of D, which lies on the incircle of ABC. We can prove that the locus is just 1/3 of the incircle.

Let $EF$ meet $OD$ at $P$. Let $EF$ meet $BO$ at $L$. Let $AO$ meet $AC$ at $M$ and the circle at $N$. Suppose $D$ is on the small arc of $AN$. Obviously, $DF \parallel NL$. So $\angle NLF + \angle DFE = 180^\circ$. Since $F,D,E,C$ is cyclic, $\angle DFE + \angle DCE = 180^\circ$. Since $B,N,D,C$ is cyclic, $\angle DCE = \angle BND$. So we have $\angle LND = \angle NLF$ and $LN \parallel DF$. So $DFLN$ is isosceles trapezoid. We have $\angle DLN = \angle LNF$. Since $M$ is the midpoint of $[ON]$, $OFN$ is isoceles. $\angle LOF = \angle LNF$. Thus, we have $\angle LOF = \angle DLN$. So $DL \parallel OF$. We also know $OL \parallel DF$. So $OLDF$ is a parallelogram. So $OP=PD = \frac {OD}2 = \frac {ON}2 = OM$. So $P$ is on the circle with center O and passing through the midpoint of $AC$. This is the incircle of $\triangle ABC$. Because of the problem statement, $D$ varies on the small arc $AC$. So the locus is an arc with measure $120^\circ$. Its endpoints are the points where $[OC]$ and $[OA]$ meet the incircle.