Congruent Triangles Worksheets - Math Monks - Free Printable

Problem 1: Prove that \( \triangle DEA \cong \triangle BCE \)

Given:
- \( \overline{DE} \cong \overline{EC} \)
- \( \overline{AE} \cong \overline{EB} \)

To Prove:
- \( \triangle DEA \cong \triangle BCE \)

Proof:
1. Given: \( \overline{DE} \cong \overline{EC} \)
2. Given: \( \overline{AE} \cong \overline{EB} \)
3. Reflexive Property of Congruence: \( \overline{EA} \cong \overline{EB} \) (since \( E \) is the same point in both triangles)
4. SSS Congruence Criterion: Since \( \overline{DE} \cong \overline{EC} \), \( \overline{AE} \cong \overline{EB} \), and \( \overline{EA} \cong \overline{EB} \), we can conclude that \( \triangle DEA \cong \triangle BCE \).

Conclusion:
\[ \boxed{\triangle DEA \cong \triangle BCE} \]

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Problem 2: Prove that \( \triangle GJH \cong \triangle HJI \)

Given:
- \( \overline{GH} \parallel \overline{JI} \)
- \( \overline{GH} \cong \overline{JI} \)

To Prove:
- \( \triangle GJH \cong \triangle HJI \)

Proof:
1. Given: \( \overline{GH} \parallel \overline{JI} \)
2. Given: \( \overline{GH} \cong \overline{JI} \)
3. Alternate Interior Angles Theorem: Since \( \overline{GH} \parallel \overline{JI} \), \( \angle GHJ \cong \angle HIJ \) (alternate interior angles).
4. Reflexive Property of Congruence: \( \overline{HJ} \cong \overline{HJ} \) (common side)
5. SAS Congruence Criterion: Since \( \overline{GH} \cong \overline{JI} \), \( \angle GHJ \cong \angle HIJ \), and \( \overline{HJ} \cong \overline{HJ} \), we can conclude that \( \triangle GJH \cong \triangle HJI \).

Conclusion:
\[ \boxed{\triangle GJH \cong \triangle HJI} \]

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Problem 3: Prove that \( \triangle WNX \cong \triangle WYX \)

Given:
- \( \angle XYP \cong \angle XNP \)
- \( \angle NWX \cong \angle YWX \)

To Prove:
- \( \triangle WNX \cong \triangle WYX \)

Proof:
1. Given: \( \angle XYP \cong \angle XNP \)
2. Given: \( \angle NWX \cong \angle YWX \)
3. Reflexive Property of Congruence: \( \overline{WX} \cong \overline{WX} \) (common side)
4. AAS Congruence Criterion: Since \( \angle NWX \cong \angle YWX \), \( \angle XYP \cong \angle XNP \), and \( \overline{WX} \cong \overline{WX} \), we can conclude that \( \triangle WNX \cong \triangle WYX \).

Conclusion:
\[ \boxed{\triangle WNX \cong \triangle WYX} \]

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Problem 4: Prove that \( \triangle BDC \cong \triangle BDE \)

Given:
- \( \overline{BD} \) bisects \( \angle B \)
- \( \overline{BD} \perp \overline{CE} \)

To Prove:
- \( \triangle BDC \cong \triangle BDE \)

Proof:
1. Given: \( \overline{BD} \) bisects \( \angle B \)
- This means \( \angle CBD \cong \angle DBE \).
2. Given: \( \overline{BD} \perp \overline{CE} \)
- This means \( \angle BDC \) and \( \angle BDE \) are right angles, so \( \angle BDC \cong \angle BDE \).
3. Reflexive Property of Congruence: \( \overline{BD} \cong \overline{BD} \) (common side)
4. ASA Congruence Criterion: Since \( \angle CBD \cong \angle DBE \), \( \overline{BD} \cong \overline{BD} \), and \( \angle BDC \cong \angle BDE \), we can conclude that \( \triangle BDC \cong \triangle BDE \).

Conclusion:
\[ \boxed{\triangle BDC \cong \triangle BDE} \]

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Problem 5: Prove that \( \angle A \cong \angle E \)

Given:
- \( \overline{BD} \perp \overline{AB} \)
- \( \overline{BD} \perp \overline{DE} \)
- \( \overline{BC} \cong \overline{DC} \)

To Prove:
- \( \angle A \cong \angle E \)

Proof:
1. Given: \( \overline{BD} \perp \overline{AB} \)
- This means \( \angle ABD \) is a right angle.
2. Given: \( \overline{BD} \perp \overline{DE} \)
- This means \( \angle BDE \) is a right angle.
3. Given: \( \overline{BC} \cong \overline{DC} \)
4. Reflexive Property of Congruence: \( \overline{BD} \cong \overline{BD} \) (common side)
5. HL Congruence Criterion for Right Triangles: Since \( \overline{BC} \cong \overline{DC} \) and \( \overline{BD} \cong \overline{BD} \), we can conclude that \( \triangle ABC \cong \triangle EBD \).
6. Corresponding Parts of Congruent Triangles are Congruent (CPCTC): Since \( \triangle ABC \cong \triangle EBD \), \( \angle A \cong \angle E \).

Conclusion:
\[ \boxed{\angle A \cong \angle E} \]

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Problem 6: Prove that \( \triangle XYZ \cong \triangle PLZ \)

Given:
- \( \angle Y \cong \angle P \)
- \( Z \) is the midpoint of \( \overline{YP} \)

To Prove:
- \( \triangle XYZ \cong \triangle PLZ \)

Proof:
1. Given: \( \angle Y \cong \angle P \)
2. Given: \( Z \) is the midpoint of \( \overline{YP} \)
- This means \( \overline{YZ} \cong \overline{PZ} \).
3. Reflexive Property of Congruence: \( \overline{ZL} \cong \overline{ZL} \) (common side)
4. SAS Congruence Criterion: Since \( \angle Y \cong \angle P \), \( \overline{YZ} \cong \overline{PZ} \), and \( \overline{ZL} \cong \overline{ZL} \), we can conclude that \( \triangle XYZ \cong \triangle PLZ \).

Conclusion:
\[ \boxed{\triangle XYZ \cong \triangle PLZ} \]

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Final Answers:

1. \( \boxed{\triangle DEA \cong \triangle BCE} \)
2. \( \boxed{\triangle GJH \cong \triangle HJI} \)
3. \( \boxed{\triangle WNX \cong \triangle WYX} \)
4. \( \boxed{\triangle BDC \cong \triangle BDE} \)
5. \( \boxed{\angle A \cong \angle E} \)
6. \( \boxed{\triangle XYZ \cong \triangle PLZ} \)