1. Given Setup and Definitions:
- Let $B, C, D, E$ be four pairwise distinct collinear points on a line.
- Let $A$ be a point not on the line $BC$.
- The circumcircle of $\triangle ABC$ meets $AD$ and $AE$ again at $F$ and $G$ respectively.
- We need to show that $DEFG$ is cyclic if and only if $AB = AC$.

2. Angle Chasing:
- Without loss of generality, assume $B, C, D, E$ are collinear in that order.
- Let $\angle FBC = x$, $\angle ABC = B$, and $\angle ACB = C$.
- Since $F$ lies on the circumcircle of $\triangle ABC$, $\angle ADF = \angle ACB = C$.
- Similarly, $\angle AEG = \angle ABC = B$.

3. Angles in Cyclic Quadrilateral:
- To show $DEFG$ is cyclic, we need to show that $\angle EDF + \angle FGE = 180^\circ$.
- $\angle EDF = \angle ADF = C$.
- $\angle FGE = \angle AGE = B$.

4. **If $AB = AC$:**
- If $AB = AC$, then $\triangle ABC$ is isosceles with $B = C$.
- Therefore, $\angle EDF = \angle FGE = B$.
- Hence, $\angle EDF + \angle FGE = B + B = 180^\circ$.
- Thus, $DEFG$ is cyclic.

5. **If $DEFG$ is Cyclic:**
- If $DEFG$ is cyclic, then $\angle EDF + \angle FGE = 180^\circ$.
- From the previous angle chasing, $\angle EDF = C$ and $\angle FGE = B$.
- Therefore, $C + B = 180^\circ$.
- Since $B$ and $C$ are angles in $\triangle ABC$, and $\angle ABC + \angle ACB + \angle BAC = 180^\circ$, it follows that $B = C$.
- Hence, $AB = AC$.

Thus, we have shown that $DEFG$ is cyclic if and only if $AB = AC$.

$\blacksquare$