, 2021

Let $\angle(p, q)$ denote the directed angle between lines $p$ and $q$.
The points $B, C, I$, and $I_A$ lie on the circle $\Gamma$ with diameter $II_A$. Let $\omega$ and $\Omega$ denote the circles $(IAEF)$ and $(AID)$, respectively. Let $T$ be the second intersection point of $\omega$ and $\Gamma$. Then $T$ is the Miquel point of the complete quadrilateral formed by the lines $BC, BI_A, CI_A$, and $DEF$, so $T$ also lies on circle $(BDE)$ (as well as on circle $(CDF)$). We claim that $T$ is the desired tangency point of $\omega$ and $\Omega$.
In order to show that $T$ lies on $\Omega$, use cyclic quadrilaterals $BDET$ and $BII_AT$ to write
$$\angle(DT, DA) = \angle(DT, DE) = \angle(BT, BE) = \angle(BT, BI_A) = \angle(IT, II_A) = \angle(IT, IA).$$
To show that $\omega$ and $\Omega$ are tangent at $T$, let $\ell$ be the tangent to $\omega$ at $T$, so that
$$\angle(TI_A, \ell) = \angle(EI_A, ET).$$
Using circles $(BDET)$ and $(BICI_A)$, we get
$$\angle(EI_A, ET) = \angle(EB, ET) = \angle(DB, DT).$$
Therefore,
$$\angle(TI, \ell) = 90^\circ + \angle(TI_A, \ell) = 90^\circ + \angle(DB, DT) = \angle(DI, DT),$$
which shows that $\ell$ is tangent to $\Omega$ at $T$.

We use the notation of circles $\Gamma, \omega$, and $\Omega$ as in the previous solution.
Let $L$ be the point opposite to $I$ in circle $\Omega$. Then $\angle IAL = \angle IDL = 90^\circ$, which means that $L$ is the foot of the external bisector of $\angle A$ in triangle $ABC$. Let $LI$ cross $\Gamma$ again at $M$.
Let $T$ be the foot of the perpendicular from $I$ onto $I_A L$. Then $T$ is the second intersection point of $\Omega$ and $\Gamma$. We will show that $T$ is the desired tangency point.

$$\angle(LT, LM) = \angle(AT, AI) \quad \text{and} \quad \angle(MT, ML) = \angle(MT, MI) = \angle(I_A T, I_A I),$$
which shows that triangles $TML$ and $TIA_A$ are similar and equioriented. So there exists a rotational homothety $\tau$ mapping $TML$ to $TIA_A$.
Since $\angle(ML, LD) = \angle(AI, AD)$, we get $\tau(BC) = AD$. Next, since
$$\angle(MB, ML) = \angle(MB, MI) = \angle(I_A B, I_A I) = \angle(I_A E, I_A A),$$
we get $\tau(B) = E$. Similarly $\tau(C) = F$. Since the points $M, B, C$, and $T$ are concyclic, so are their $\tau$-images, which means that $T$ lies on $\omega = \tau(\Gamma)$.
Finally, since $\tau(L) = A$ and $\tau(B) = E$, triangles $ATL$ and $ETB$ are similar so that
$$\angle(AT, AL) = \angle(ET, EB) = \angle(EI_A, ET).$$
This means that the tangents to $\Omega$ and $\omega$ at $T$ make the same angle with the line $I_A TL$, so the circles are indeed tangent at $T$.