Let: $\ell$ denote the Gauss line ; $X,Y,Z$ be the midpoints of segments $AC,BD,EF$, respectively ; $\Phi,\Psi$ be inversion wrt $\odot(O),\odot(EF)$, respectively ; $E^* = \Phi(E),F^* = \Phi(F)$ ; $\Omega = \odot(OI), \omega_1 = \odot(ENQ), \omega_2 = \odot(FMP)$. Note $E^*,F^* \in \Omega$ by Brokards. Claim 1: $\Psi$ fixes $\odot(O),\Omega,\omega_1,\omega_2$. Proof: As $\Phi$ fixes $\odot(EF)$ (as $E^*,F^* \in \odot(EF)$ by Brokards), so $\odot(O), \odot(EF)$ are orthogonal, in particular $\Psi$ fixes $(O)$. By IMOSL 2008 G4, $\Psi$ swaps $X,Y$. Now by Desargues Involution Theorem on quadrilateral $ABCD$ and line $\ell$, there is a involution swapping pairs $\{N,Q\},\{M,P\},\{X,Y\},\{\ell \cap \odot(O)\}$. Since $\Psi$ swaps the first two pairs, it also must swap all fours, which implies our claim. $\square$ Claim 2: $F^* \in \omega_1$ and $E^* \in \omega_2$ Proof: It is well known that $F^*$ is the center of spiral similarity $\tau$ sending $\overline{DA} \to \overline{BC}$. Since $F^* \in \Omega$, so by converse of generalized IMO 2005/5, $\tau$ also sends $Q \to N$, thus $F^* \in \omega_1$.$\square$ By Claim 1, power of $Z$ wrt both of $\Omega$ and $\omega_1$ is equal to $(ZF^*)^2$ (as $F^* \in \odot(EF)$), which implies $\overline{ZF^*}$ is tangent to both $\Omega,\omega_1$. In particular, $\Omega$ and $\omega_1$ are tangent to each other at $F^*$, completing the proof. $\blacksquare$

Beautiful problem! Let $X,Y,Z$ be midpoints of $AC,BD,EF$ respectively. Note $\triangle EBD \sim \triangle EAC \implies \triangle EAX \sim \triangle EBY$ so $X,Y$ are isogonal wrt $\angle DEC.$ Similarly, $X,Y$ are isogonal wrt $\angle DFA.$ So \begin{align*} \measuredangle EXF + \measuredangle EYF &= (\measuredangle EDF + \measuredangle DFX + \measuredangle XED) + (\measuredangle EDF + \measuredangle DFY + \measuredangle YED) \\ &= 2\measuredangle EDF + \measuredangle DEC + \measuredangle DFA \\ &= 180^\circ \end{align*}and $ZX \cdot ZY = ZF^2.$ So $\triangle ZFX \sim \triangle ZYF \implies \measuredangle ZYF = \measuredangle XFZ.$ But $\measuredangle PFZ = \measuredangle ZYF - \measuredangle PFY = \measuredangle XFZ - \measuredangle XFM = \measuredangle MFZ.$ So $\triangle PFZ \sim \triangle FMZ$ and $ZF^2 = ZM\cdot ZP.$ Let $K$ be the miquel point of $ABCD.$ Note $\measuredangle AKO = 90^\circ - \measuredangle EKA = 90^\circ - \measuredangle CDA = \measuredangle OAI \implies \triangle OAI \sim OKA$ so $OA^2 = OI \cdot OK.$ Since $OK \perp EF,$ the inversion wrt $\odot(O)$ swaps $EF$ and $\odot(OI).$ Let $E'$ be image of $E$ after inversion wrt $\odot(O).$ So $E'$ is on $\odot(OI).$ It is well known that $\odot(O)$ and $\odot(EF)$ are orthogonal, so $ZE' = ZF.$ Note $(AEKB), (DEKC)$ map to $(AE'IB), (DE'IC)$ so $AE'IB$ and $DE'IC$ are cyclic and a spiral similarity centered at $E'$ takes $AB$ to $CD.$ By Menelaus on $\triangle ABC, \triangle DBC,$ we get $\frac{AM}{MB} = \frac{CN}{BN} = \frac{CP}{PD}$ so the spiral similarity takes $M$ to $P.$ So $\angle E'PF = \angle E'PD = \angle E'MF$ and $E'$ is on $(FMP).$ So $ZE'$ is tangent to $\odot(O)$ and $\odot(EF)$ at $E'$ and with a symmetrical argument on $(ENQ)$ we're done. $\blacksquare$