1. **State the problem:** Given that $\angle G \cong \angle I$ and that $FH$ bisects $\angle GFI$, prove that $\triangle GFH \cong \triangle IFH$. 2. **Write down the given information and what to prove:** - Given: $\angle G \cong \angle I$ and $FH$ bisects $\angle GFI$. - To prove: $\triangle GFH \cong \triangle IFH$. 3. **Recall definitions and properties:** - Definition of angle bisector: $FH$ bisects $\angle GFI$ means $\angle GFH \cong \angle IFH$. - Reflexive property: segment $FH$ is congruent to itself. 4. **Fill in the proof table:** - Statement 1: $\angle G \cong \angle I$ and $FH$ bisects $\angle GFI$ (Given). - Statement 2: $\angle GFH \cong \angle IFH$ (Definition of angle bisector). - Statement 3: $FH \cong FH$ (Reflexive property). 5. **Use the Angle-Side-Angle (ASA) congruence postulate:** - In $\triangle GFH$ and $\triangle IFH$: - $\angle G \cong \angle I$ (Given), - $FH \cong FH$ (Reflexive), - $\angle GFH \cong \angle IFH$ (Angle bisector definition). 6. **Conclusion:** - By ASA postulate, $\triangle GFH \cong \triangle IFH$. **Final answer:** $\triangle GFH \cong \triangle IFH$