Triangle Congruence Proofs Practice worksheet featuring three geometry problems with diagrams and tables for proving triangle congruence using given information.

Here's the complete solution for all three triangle congruence proofs, with explanations.

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## Problem 1

Given: C is the midpoint of $\overline{BE}$ and $\overline{AD}$.
Prove: $\triangle ABC \cong \triangle DEC$

Completed Proof Table:

| Statement | Reason |
|--------------------------------------|-------------------------------|
| 1. C is midpoint of $\overline{BE}$ and $\overline{AD}$ | 1. Given |
| 2. $\overline{BC} \cong \overline{EC}$ | 2. Definition of a midpoint |
| 3. $\overline{AC} \cong \overline{DC}$ | 3. Definition of a midpoint |
| 4. $\angle ACB \cong \angle DCE$ | 4. Vertical angles are congruent |
| 5. $\triangle ABC \cong \triangle DEC$ | 5. SAS Congruence Postulate |

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✔ Explanation:

- Since C is the midpoint of $\overline{BE}$, it divides it into two equal parts → $\overline{BC} \cong \overline{EC}$.
- Similarly, since C is the midpoint of $\overline{AD}$, → $\overline{AC} \cong \overline{DC}$.
- The angles at point C ($\angle ACB$ and $\angle DCE$) are vertical angles — they are formed by intersecting lines and are always congruent.
- So we have: two sides and the included angle congruent → SAS (Side-Angle-Side) congruence.

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## Problem 2

Given: $\overline{BC} \cong \overline{DA}$ and $\overline{AC}$ bisects $\angle BCD$
Prove: $\triangle ABC \cong \triangle CDA$

*(Note: The diagram suggests quadrilateral ABCD with diagonal AC.)*

Completed Proof Table:

| Statement | Reason |
|--------------------------------------|-------------------------------|
| 1. $\overline{BC} \cong \overline{DA}$ | 1. Given |
| 2. $\overline{AC}$ bisects $\angle BCD$ | 2. Given |
| 3. $\angle BCA \cong \angle DCA$ | 3. Definition of angle bisector |
| 4. $\overline{AC} \cong \overline{AC}$ | 4. Reflexive Property |
| 5. $\triangle ABC \cong \triangle CDA$ | 5. SAS Congruence Postulate |

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✔ Explanation:

- We’re given that side BC ≅ DA.
- Since AC bisects ∠BCD, it splits it into two equal angles → ∠BCA ≅ ∠DCA.
- Side AC is common to both triangles → reflexive property: AC ≅ AC.
- So again, we have two sides and the included angle congruent → SAS.

> 💡 Note: The order of vertices matters. Triangle ABC corresponds to triangle CDA because:
> - B ↔ D (given sides BC ≅ DA)
> - C ↔ C (shared vertex)
> - A ↔ A (shared vertex)

So the correspondence is correct.

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## Problem 3

Given: $\angle X \cong \angle H$ and $\overline{XG} \parallel \overline{FH}$
Prove: $\triangle XGF \cong \triangle HFG$

*(Note: Points X, G, F, H form a quadrilateral with diagonal GF.)*

Completed Proof Table:

| Statement | Reason |
|--------------------------------------|-------------------------------|
| 1. $\angle X \cong \angle H$ | 1. Given |
| 2. $\overline{XG} \parallel \overline{FH}$ | 2. Given |
| 3. $\angle FGX \cong \angle GFH$ | 3. Alternate Interior Angles Theorem (since XG ∥ FH and GF is transversal) |
| 4. $\overline{GF} \cong \overline{GF}$ | 4. Reflexive Property |
| 5. $\triangle XGF \cong \triangle HFG$ | 5. AAS Congruence Theorem |

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✔ Explanation:

- We’re given ∠X ≅ ∠H.
- Since XG ∥ FH and line GF crosses them, then ∠FGX and ∠GFH are alternate interior angles → they are congruent.
- Side GF is shared → reflexive property.
- So we have: two angles and a non-included side congruent → AAS (Angle-Angle-Side).

> 💡 Why not ASA? Because the side is NOT between the two angles — in triangle XGF, angles are at X and G, and side GF is opposite angle X. So it’s AAS.

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## ✔ Final Summary:

| Problem | Congruence Method Used |
|---------|------------------------|
| 1 | SAS |
| 2 | SAS |
| 3 | AAS |

These are standard triangle congruence postulates used in geometry proofs.

Let me know if you’d like diagrams or further clarification!