Congruent Triangles Proof Worksheet | PDF | Triangle Geometry ... - Free Printable

Explanation:
Let’s solve each problem step by step, filling in the statements and reasons as required.

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Problem 1:
Given: $\overline{BC} \cong \overline{DC}$; $\overline{AC} \cong \overline{EC}$
Prove: $\triangle BCA \cong \triangle DCE$

We need to fill in the proof table:

| Statements | Reasons |
|------------|---------|
| 1. $\overline{BC} \cong \overline{DC}$, $\overline{AC} \cong \overline{EC}$ | 1. Given |
| 2. $\angle BCA \cong \angle DCE$ | 2. Vertical ∠s Theorem (angles formed by intersecting lines at point C are vertical angles) |
| 3. $\triangle BCA \cong \triangle DCE$ | 3. SAS (Side-Angle-Side) Congruence Postulate — two sides and the included angle are congruent |

So blank #3 is SAS.

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Problem 2:
Given: $\overline{JK} \cong \overline{LK}$; $\overline{JM} \cong \overline{LM}$
Prove: $\triangle KJM \cong \triangle KLM$

Let’s list what we know:
- $\overline{JK} \cong \overline{LK}$ (given)
- $\overline{JM} \cong \overline{LM}$ (given)
- $\overline{KM} \cong \overline{KM}$ (common side — reflexive property)

So the proof should be:

| Statements | Reasons |
|------------|---------|
| 1. $\overline{JK} \cong \overline{LK}$, $\overline{JM} \cong \overline{LM}$ | 1. Given |
| 2. $\overline{KM} \cong \overline{KM}$ | 2. Reflexive Property |
| 3. $\triangle KJM \cong \triangle KLM$ | 3. SSS (Side-Side-Side) Congruence Postulate |

So blank #3 is SSS.

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Problem 3:
Given: $\angle G \cong \angle I$; $\overline{FH}$ bisects $\angle GFI$
Prove: $\triangle GFH \cong \triangle IFH$

Let’s unpack:
- $\angle G \cong \angle I$ (given)
- $\overline{FH}$ bisects $\angle GFI$ → means $\angle GFH \cong \angle IFH$ (definition of angle bisector)
- $\overline{FH} \cong \overline{FH}$ (reflexive property — common side)

Now we have:
- $\angle G \cong \angle I$ (angle)
- $\angle GFH \cong \angle IFH$ (angle)
- $\overline{FH} \cong \overline{FH}$ (side between the two angles)

That’s AAS (Angle-Angle-Side), or could also be ASA if the side were included — but here the side $\overline{FH}$ is *between* the two angles? Let's check:

In $\triangle GFH$: angles are $\angle G$, $\angle GFH$, and side $\overline{FH}$ is between $\angle GFH$ and $\angle GHF$. Wait — better to label carefully:

Vertices:
- Triangle GFH: angles at G, F, H
- Triangle IFH: angles at I, F, H

We know:
- $\angle G \cong \angle I$ (angle at G and I)
- $\angle GFH \cong \angle IFH$ (angle at F in both triangles — since FH bisects ∠GFI)
- Side $\overline{FH}$ is common — it's opposite angle G in triangle GFH, and opposite angle I in triangle IFH.

So we have two angles and a non-included side — that’s AAS, which is a valid congruence theorem.

Alternatively, some textbooks accept AAS as a corollary of ASA.

But let’s fill the table as instructed:

| Statements | Reasons |
|------------|---------|
| 1. $\angle G \cong \angle I$; $\overline{FH}$ bisects $\angle GFI$ | 1. Given |
| 2. $\angle GFH \cong \angle IFH$ | 2. Def. of angle bisector |
| 3. $\overline{FH} \cong \overline{FH}$ | 3. Reflexive Prop. |
| 4. $\triangle GFH \cong \triangle IFH$ | 4. AAS |

So blank #2: “angle bisector”
Blank #4: AAS

Let me double-check: Yes — two angles and a non-included side → AAS is correct.

Some might argue ASA if we consider $\angle GHF \cong \angle IHF$, but we don’t know that — we only know the two angles at G/I and at F. So AAS is safest and correct.

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Final answers per problem:

1. Statement 3 reason: SAS
2. Statement 3: SSS
3. Statement 2 reason: angle bisector
Statement 4 reason: AAS

Now write the final answer in required format.

Final Answer:
1. SAS
2. SSS
3. angle bisector; AAS