It's easy to see that $ID = IE = IF$. Hence $IM = IN$. But $AI$ is bisector of $\angle{BAC}$ then $I$ $\in$ $(AMN)$. So $\angle{MIN} = 180^{\circ} - \angle{BAC} = \angle{EIF}$. Hence $\triangle IEF = \triangle IMN$ or $EF = MN,$ which means $MENF$ is isosceles trapezoid. Therefore $EN$ $\parallel$ $FM$. Note that $\angle{NME} = \angle{EFN}$ then $\triangle AEF = \triangle ANM$ or $AE = AN, AM = AF$. Hence $AI$ $\perp$ $EN$. Since $IP \cdot IB = ID^2 = IQ \cdot IC$ then $B, P, Q, C$ lie on a circle. So $AI$ $\perp$ $PQ$ or $EN$ $\parallel$ $FM$ $\parallel$ $AI$. We also see that the $I$ - median of $\triangle IPQ$ is $I$ - symmedian of $\triangle IBC,$ which passes through midpoint of arc $BAC$

Claim 1 : $I$ is the circumcenter of $(DEF)$ Proof : easy to notice by simple angle chasing that $ID=IE=IF$ Claim 2 : $IQDP$ is cyclic Proof : $\angle{IQF}=\angle{IFC}$$(\triangle{IQF}\sim \triangle{IFC})$$=\angle{IDB}=\pi-\angle{IEB}=\angle{IPD}$$(\triangle{IEB}\sim \triangle{IPE})$ Claim 3 : $EN , FM , PQ$ are parallel Proof : $\angle{QPD}=\angle{DIQ}=\angle{DFN}=\angle{NED}$ $\implies$ $PQ \parallel EN$ And $\angle{PQD}=\angle{PID}=\angle{DEB}=\angle{MFD}$ $\implies$ $PQ \parallel MF$ Claim 4 : The $I$-median of $\triangle{IPQ}$ bisects $\overarc{BAC}$ Proof : Let $K$ be the midpoint of segment $\overline{PQ}$ , and $L$ the midpoint of segment $\overline{BC}$ By simple angle chasing $PQCB$ is cyclic , hence $\triangle{IKQ} \sim \triangle{ILB}$ , wich implies that the $I$-median of $\triangle{IPQ}$ is the $I$-symmedian of $\triangle{IBC}$.$(1)$ Now let $X$ be the intersection of the tangents to $(BIC)$ at $B$ and $C$ , angle chasing gives that $X$ $\in$ $(ABC)$ , thus it suffices to show that points $X , I , K$ are collinear , wich is obviously true by $(1)$.

Nice angle chasing exercise. $\angle PIQ+\angle PDQ=90+\angle A/2+ \angle B/2+\angle C/2=180 \implies IPDQ$ is a cyclic quadrilateral. $\angle MFD=\angle MED=\angle BID= \angle PQD \implies PQ \parallel MF$ and by symmetry $PQ \parallel EN$. $\angle IPQ=\angle IDQ=\angle ICF=\angle ICD \implies PQCD$ is a cyclic quadrilateral so $I$-median of $IPQ$ is the same as $I$-symmedian of $IBC$. Proof of the fact that $I$-symmedian of $IBC$ passes through the midpoint of arc $BAC$: $K$ is the midpoint of arc $BAC$, $BI \cap (ABC)=U$, $CI\cap (ABC)=V$, $T=KI \cap BC, I\cap BC=X, AI \cap (ABC)=S$. From angle chasing $KUIV$ is a parallelogram. We know that $IX$ and $IS$ are isogonal conjugates, $\angle TIX=\angle TKM=\angle SIM \implies IT$ and $IM$ are isogonal conjugates as desired.