Very interesting problem...Any ideas?

Let the tangent of $\odot (ABC)$ at $A$ intersect $BC$ at $G$.Let $FE\cap AG=H,FD\cap AC=I,ED\cap AB=J$.We are going to prove that $\odot (DPQ)$ pass though the fixed point $G$. Claim:$A,H,J,Q,E$ are concyclic;$A,F,P,H,I$ are concyclic. By angle chasing:$\angle FED=\angle ACB=\angle HAB$.So $A,H,J,E$ are concyclic. $\angle AQE=\angle AQC-\angle EQC=\angle ABC-\angle EDC=\angle AJE$.So $A,J,Q,E$ are concyclic. So $A,H,J,Q,E$ are concyclic.Likewise,$A,F,P,H,I$ are concyclic. Notice that:$\angle QDC=\angle QEC=\angle QHA$.So $G,H,D,Q$ are concyclic. Likewise,$G,H,D,P$ are concyclic. So we can get:$G,P,D,Q,H$ are concyclic,which implies that $G\in \odot (DPQ)$.Done.$\Box$

Lemma 1. The secend intersection of $(BDF),(CED)$ is a fix point. Proof.call that point $X$. Note that $\angle FXD = 180 -\angle B$ and $\angle EXD=180- \angle B$ So $\angle EXF =180 - \angle A$ . So $AEXF$ is cyclic. We have : $$ \angle BXC= \angle A + \angle ABX + \angle ACX =\angle A+ \angle FDX + \angle EDX = 2\angle A$$Similarly $ \angle AXB = \angle B + \angle C = \angle AXC$ So $X$ is a fix point. Also , its well-known that such $X$ , is in fact the midpoint of $A$-symmedian chord in $(ABC)$. Now we shall prove that the desired point is $T$ , the intersection of the tangent through $A$ in $(ABC)$ with $BC$. Let $O$ be the center of $(ABC)$. consider an inversion through $T$ withbradious $TA$. since $X$ was the midpoint of symmedian chord and the other tangent through $T$ to $(O)$ is the intersection of A-symmedian with $(O)$ , $X$ goes to $O$ . $B,C$ swaps and $D$ goes to another point $D'$ on $BC$. So $P,Q$ go to $(D'OC) \cap (O)$ and $(BD'O) \cap (O)$ respectively and we shall prove that in this case $D',P',Q'$ are collinear. Now apply another inversion , this time with center $O$ and radius $OA$. So $D'$ goes to some point on the circle.$(BOC)$ and $P',Q'$ would be $D''B \cap (O)$ and $D''C \cap (O)$ and we should prove that $P'',Q'',D'',O$ are concyclic. This is obvious since $OD"$ is the angle bisector of $BD"C$ and so $PD"Q$ where we also have $OP''=OQ''$ and $P''D''=Q''D''$ only occures when $BD''=CD''$ which is a trivial case and we're done

Let $O$ be the circumcenter of $ABC$, $T = AA \cap BC$, the projection of $A$ onto $OT$ be $X$, the midpoints of $AB$ and $AC$ be $M$ and $N$ respectively, and the projection of $A$ onto $BC$ be $L$. Since $X$ is the $A$-Dumpty point, it's well-known that $BXOC$ is cyclic and $(AMN), (BLM), (CNL)$ concur at $X$. Now, observe $$LMN \overset{-}{\sim} AMN \overset{+}{\sim} ABC \overset{-}{\sim} DFE$$so $LMN \overset{+}{\sim} DFE$. Hence, the Spiral Center Lemma implies $(AMN)$, $(BLM)$, $(CNL)$, $(AEF)$, $(BDF)$, and $(CDE)$ concur. It follows that $X$ is the second intersection point between $(BDF)$ and $(CDE)$. By Radical Axes, we know $BP, CQ, DX$ concur at some point $R$. Now, let $D^*$ denote the inverse of $D$ wrt $(ABC)$. It's easy to see $D^* \in (BXOC)$. Moreover, because $X$ and $T$ are clearly inverses wrt $(ABC)$ as well, we know $(BPTD^*)$ and $(CQD^*T)$ are cyclic. Thus, applying Radical Axes again yields $R \in D^*T$. Now, consider the inversion about $R$ with radius $\sqrt{RB \cdot RP}$. By POP, we have $B \leftrightarrow P$, $C \leftrightarrow Q$, $D \leftrightarrow X$, and $D^* \leftrightarrow T$. Since $BXCD^*$ is cyclic, it follows that $PDQT$ is cyclic, which suffices. $\blacksquare$ Remark: I determined that $T$ is the fixed point by considering the case where $AD \perp BC$ and performing $\sqrt{bc}$-inversion.

Take foot of perpendicular from $A$ to $BC$ be $D'$, midpoint of $AB$ and $BC$ be $F'$ and $E'$,respectively. Let $T$ be the $A-$Dumpty point. Take $O$ as the circumcenter of $(ABC)$ Notice that $BF'D'T$, $CE'D'T$ and $AE'F'T$ are cyclic and $D'E'F'$ is similar to $DEF$. Note that the spiral homothety center that sends $DEF$ to $D'E'F'$ must be on circle $BF'D'T$ and $CE'D'T$ so it must be $T$. Also converse of this is also true: if you take any $D$ on $BC$, $E$ and $F$ are defined as the intersections of $(TDC)$ and $(TDB)$ with $AC$ and $AB$ Rewrite the problem as follow: $T$ is the $A$-Dumpty point of $ABC$. $D$ is taken on $BC$ and $(TDB)$ and $(TDC)$ intersect $(ABC)$ at $P$ and $Q$. Show that $(DPQ)$ passes through a fixed point. If you take inversion with center $T$ problem turns to this: $D$ is taken on $(BOC)$, $BD$ and $CD$ intersect circumcircle at $P$ and $Q$. Show that $(DPQ)$ passes through a fixed point. Indeed, that fixed point is $O$ in the inverted picture. In the original picture, it is $OT\cap BC$ or it can be defined as the intersection of tangent from $A$ with $BC$.

Suppose that $S$ is Miquel point of $D, E, F$ with respect to $\triangle ABC$. Since $\triangle DFE \sim \triangle ABC,$ we have $\angle{DFE} = \angle{ABC}$. But $\angle{DBS} = \angle{DFS}$ then $\angle{SBF} = \angle{SFE} = \angle{SAE},$ which means $AC$ touches $(ABS)$. Similarly, $AB$ touches $(ACS)$. So $S$ is $A$ - Dumpty point of $\triangle ABC$. Let $S'$ be inversion of $S$ through $(ABC),$ $O$ be circumcenter of $\triangle ABC$. We have $$\angle{SFE} = \angle{SAE} = \angle{SBA} = \angle{SDF}$$Similarly, $\angle{SEF} = \angle{SDE}$. Hence $EF$ is common tangent of $(DSF), (DSE)$. So we have $$\angle{PS'Q} = \angle{PS'O} + \angle{QS'O} = \angle{OPS} + \angle{OQS} = \angle{POQ} - \angle{PSQ}$$$$= \angle{POQ} - \angle{PSD} - \angle{DSQ} = \angle{POQ} - \angle{PFE} - \angle{FEQ} + \angle{EFD} + \angle{DEF}$$$$= \angle{POQ} + \angle{PDF} + \angle{QDE} - 180^{\circ} - \angle{EDF} = \angle{POQ} + \angle{ACP} + \angle{ACQ} - 180^{\circ} - \angle{EDF}$$$$= 360^{\circ} - \angle{ACP} - \angle{ACQ} - 180^{\circ} - \angle{EDF} = 180^{\circ} - \angle{PDF} - \angle{PDE} - \angle{EDF} = 180^{\circ} - \angle{PDQ}$$or $P, Q, D, S'$ lie on a circle