Let me analyze the given diagram and solve the problem.The problem requires proving $\triangle DFG \cong \triangle EFG$ based on the given information.

Given:

1. $\overline{FG} \perp \overline{DE}$ (indicating that $FG$ is perpendicular to $DE$).
2. $G$ is the midpoint of $\overline{DE}$.

To Prove:

$\triangle DFG \cong \triangle EFG$.

Proof:

1. Midpoint Definition: $G$ is the midpoint of $\overline{DE}$, so: $\overline{DG} \cong \overline{GE}.$

2. Perpendicular Bisector: $\overline{FG} \perp \overline{DE}$, meaning: $\angle DGF = \angle EGF = 90^\circ.$

3. Shared Side: $\overline{FG}$ is common to both $\triangle DFG$ and $\triangle EFG$, so: $\overline{FG} \cong \overline{FG}.$

4. Conclusion (RHS Congruence): By the RHS (Right Angle-Hypotenuse-Side) Congruence Theorem, $\triangle DFG \cong \triangle EFG$ because:

   • Both triangles have a right angle ($\angle DGF = \angle EGF = 90^\circ$),
   • Their hypotenuses are congruent ($\overline{DG} \cong \overline{GE}$),
   • They share a common side (( \overline{FG} \