My solution: Since $ BC $ is the image of $ \odot (ABC) $ under Inversion $ \mathbf{I}(A, AB) = \mathbf{I}(A, AC)=\Psi $ , so $ D \longleftrightarrow F, E \longleftrightarrow AE \cap BC $ under $ \Psi \Longrightarrow G \in BC \Longrightarrow AG \cdot AE=AC^2 \Longrightarrow AC $ is tangent to $ \odot (CEG) $ at $ C $ . Q.E.D

My sol is similar to TelvCohl, but my last step: Let $AC$ intersect $\odot (CEG)$ at $C'$. Then, as the inverse of $\odot (CEG)$ under $\Psi$ is $\odot (CEG)$ itself and as $C$ lies on $\odot (CEG)$, it follows that the inverse of $C$ under $\Psi$ is $C'$ or the inverse of $C'$ under $\Psi$ is $C'$ itself. In either case, $C'$ coincides with $C$.

Any synthetic solution, not including inversion?

G is actually on BC. You can prove it with angle chasing. Then some more to finish the problem.

navredras wrote: Any synthetic solution, not including inversion? Solution: $ ABC=ACB=BDF$ and $ABC+DBA=BDF+AFC=ABC+AFC $ so $ AFB=ABD $ and we get that $ AB $ is tangent to circumcircle of $\triangle BDF $ so $ AB^2=AD \cdot AF=AC^2=AE \cdot AG $ DONE

Well, this is pretty much the same solution as TelvCohl's and andria's ones. By inversion, $AD \cdot AF = AB^2$. From the power of A : $ AD \cdot AF=AE \cdot AG$. Since $AB=AC$, we obtain $AC^2=AE \cdot AG$, so $AC$ is tangent to $\odot CEG$.

To see $G$ lies on $BC$, let $ AE \cap BC = \{K\} $. $\angle ADE = \angle ACE = \alpha$, $\angle CAE = \beta \Longrightarrow \angle ABC = \angle ACB = \alpha + \beta$. Hence $\angle AKF = \alpha + \beta - \beta = \alpha$. $\angle AKF = \angle ADE = \alpha$, which means $FDEK$ is cyclic so $K$ coincides with $G$. Since $\triangle ACE \sim \triangle AGC$ from $AA$ similarity, we get $|AC|^2 = |AE| \cdot |AG|$. This finishes the proof.

I have a proof by contradiction.......We are supposed to prove that $|AC|^2 = |AE|\cdot|AG|$ or that is we have to prove $|AB|^2 = |AD|\cdot|AF|$ which is equivalent to prove that $\angle DBA = \angle BFA = \angle DEA$.....or it trivially follows to prove that $G$ lies on the line $BC$.....assume the contrary then apply Pascal to the degenerate hexagon namely $FDEGCB$ to get $ FC \cap BD $, $ DG \cap EC $ and $ FG \cap EB $ to be collinear and thus we arrive at a contradiction.... Q.E.D

$\angle ABC = \angle ACB = \angle FDB$ so $AE \cdot AG = AD \cdot AF = AB^2 = AC^2$

Nice solution !

Just Posting for storage.... Turkey TST 2015 wrote: Let $ABC$ be a triangle such that $|AB|=|AC|$ and let $D,E$ be points on the minor arcs $\overarc{AB}$ and $\overarc{AC}$ respectively. The lines $AD$ and $BC$ intersect at $F$ and the line $AE$ intersects the circumcircle of $\triangle FDE$ a second time at $G$. Prove that the line $AC$ is tangent to the circumcircle of $\triangle ECG$. Solution:- Construct a circle $\omega$ with raduis $AB$ centered at $A$, obviously $C\in\omega$. So, $\Psi_{\omega}:(\odot(ABC))\mapsto BC$. Hence, $\Psi_{\omega}:D\mapsto F$. So, $AD.AF=AE.AG=AB^2=AC^2\implies AC$ is tangent to $\odot(ECG)$.

aloski1687 wrote: Let $ABC$ be a triangle such that $|AB|=|AC|$ and let $D,E$ be points on the minor arcs $\overarc{AB}$ and $\overarc{AC}$ respectively. The lines $AD$ and $BC$ intersect at $F$ and the line $AE$ intersects the circumcircle of $\triangle FDE$ a second time at $G$. Prove that the line $AC$ is tangent to the circumcircle of $\triangle ECG$. Clearly by angle Chasing $AB$ is tangent to $\odot{BDF}$.Hence \[AE\cdot AG=AD\cdot AF=AB^2=AC^2 \implies \odot{ECG} \text{tangent to} AC\].

This image is highly symmetric One can argue that F, B and G are collinear and easy angle chase shows that AB is tangent to DBF, then since the diagram is symmetric, AC is tangent to ECG

So first of all let me define an inversion $\psi(A,\sqrt{AB.AC})=\psi(A,AB)$ By this inversion point $D$ maps to point $F$, by the power of point we have that $AD.AF=AE.AG$. But by inversion we know that $AD.AF=AB^2=AC^2$, thus we have that $AE.AG=AC^2$. But if we watch closely we now know form this that the points $F,B,C,G$ are all collinear. Because from $AE.AG$ we got that from $E$ maps to $G$ under the inversion $\psi$ and also now we can consider the power of point $A$ with respect to the circle around $\triangle ECG$, we have that $AE.AG = AC^2$. Now this means that $AC$ is tangent to the circle......

By performing an inversion with centre $A$ and radius $AB$ we know that $AD \cdot AF = AB^2$ Now let $BC \cap \odot(FDE) = G'$ $$\measuredangle DEG' = \measuredangle DFG' = \measuredangle DFB = \measuredangle DBA = \measuredangle DEA \implies A-E-G' \implies G' \equiv G$$Now $$AD \cdot AF = AB^2 = AC^2 = AE \cdot AG \ \blacksquare$$

Let $\angle ABC=\angle ACB=\angle FDB$ Let $\angle DFB=\angle ABD$ $\implies \triangle ADB \simeq \triangle ABF$ $\implies AD.AF=AB^2$ but $AB^2=AC^2$ by $FDEQ$: $AD.AF=AE.AG$ Then: $AD.AF=AB^2=BC^2=AE.AG$ this means that $AC$ is tangent to the circle $\odot{ECG}$