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To solve the problem, we need to determine the congruence of triangles in each given diagram. We will use the standard triangle congruence criteria: Side-Side-Side (SSS), Side-Angle-Side (SAS), Angle-Side-Angle (ASA), Angle-Angle-Side (AAS), and Hypotenuse-Leg (HL) for right triangles.

Let's analyze each diagram step by step:

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Diagram 13:

- Triangles: \( \Delta FIG \) and \( \Delta HIG \)
- Observations:
- \( FI = HI \) (given as marked sides)
- \( IG \) is a common side
- \( \angle FIG = \angle HIG \) (given as marked angles)
- Conclusion: By SAS (Side-Angle-Side), \( \Delta FIG \cong \Delta HIG \).

Answer for Diagram 13:
\[ \Delta FIG \cong \Delta HIG \]

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Diagram 14:

- Triangles: \( \Delta ABC \) and \( \Delta CDE \)
- Observations:
- \( AB \parallel DE \) (implied by the parallel lines)
- \( AC \parallel BE \) (implied by the parallel lines)
- \( \angle BAC = \angle ECD \) (alternate interior angles)
- \( \angle ABC = \angle CDE \) (alternate interior angles)
- \( BC \) is a common side
- Conclusion: By ASA (Angle-Side-Angle), \( \Delta ABC \cong \Delta CDE \).

Answer for Diagram 14:
\[ \Delta ABC \cong \Delta CDE \]

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Diagram 15:

- Triangles: \( \Delta AFG \) and \( \Delta AHG \)
- Observations:
- \( AF = AH \) (given as marked sides)
- \( FG = HG \) (given as marked sides)
- \( AG \) is a common side
- Conclusion: By SSS (Side-Side-Side), \( \Delta AFG \cong \Delta AHG \).

Answer for Diagram 15:
\[ \Delta AFG \cong \Delta AHG \]

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Diagram 16:

- Triangles: \( \Delta NOP \) and \( \Delta QOP \)
- Observations:
- \( NO = QO \) (given as marked sides)
- \( OP \) is a common side
- \( \angle NOP = \angle QOP \) (given as marked angles)
- Conclusion: By SAS (Side-Angle-Side), \( \Delta NOP \cong \Delta QOP \).

Answer for Diagram 16:
\[ \Delta NOP \cong \Delta QOP \]

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Diagram 17:

- Triangles: \( \Delta STU \) and \( \Delta RTU \)
- Observations:
- \( ST = RT \) (given as marked sides)
- \( TU \) is a common side
- \( \angle STU = \angle RTU \) (given as marked angles)
- Conclusion: By SAS (Side-Angle-Side), \( \Delta STU \cong \Delta RTU \).

Answer for Diagram 17:
\[ \Delta STU \cong \Delta RTU \]

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Diagram 18:

- Triangles: \( \Delta JKM \) and \( \Delta LKM \)
- Observations:
- \( JK = LK \) (given as marked sides)
- \( KM \) is a common side
- \( \angle JKM = \angle LKM \) (given as marked angles)
- Conclusion: By SAS (Side-Angle-Side), \( \Delta JKM \cong \Delta LKM \).

Answer for Diagram 18:
\[ \Delta JKM \cong \Delta LKM \]

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Final Answers:

\[
\boxed{
\begin{array}{ll}
1. \Delta FIG \cong \Delta HIG & 7. \Delta ACB \cong \Delta CDE \\
2. \Delta NOP \cong \Delta QOP & 8. \Delta GFI \cong \Delta HGI \\
3. \Delta ABC \cong \Delta CDE & 9. \Delta KLM \cong \Delta LKM \\
4. \Delta STU \cong \Delta RTU & 10. \Delta PON \cong \Delta QOP \\
5. \Delta JKM \cong \Delta LKM & 11. \Delta JKM \cong \Delta LKM \\
6. \Delta ORN \cong \Delta ORP & 12. \Delta SUR \cong \Delta TUR \\
13. \Delta FIG \cong \Delta HIG & 14. \Delta CAB \cong \Delta CDE \\
15. \Delta AFG \cong \Delta AHG & 16. \Delta NOP \cong \Delta QOP \\
17. \Delta RUS \cong \Delta TUR & 18. \Delta JKM \cong \Delta LKM \\
\end{array}
}
\]