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

Let's solve each problem step by step, following the instructions provided:
a. Show given information in the diagram using tick marks and arcs.
b. Identify other congruent parts (vertical angles, shared sides, etc.).
c. State the postulate or theorem that proves the triangles congruent.
d. Fill in the proof table.

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Problem 1

Given:
- $ \overline{BC} \cong \overline{DC} $
- $ \overline{AC} \cong \overline{EC} $

Prove: $ \triangle BCA \cong \triangle DCE $

#### Diagram Analysis (Part a & b):
- Mark $ \overline{BC} \cong \overline{DC} $ with one tick mark.
- Mark $ \overline{AC} \cong \overline{EC} $ with another tick mark.
- Notice that $ \angle ACB $ and $ \angle ECD $ are vertical angles, so they are congruent.
- Mark them with arcs to show congruence.

So we have:
- Two pairs of congruent sides
- Included angle between them congruent (by vertical angles)

This fits the SAS (Side-Angle-Side) Congruence Postulate.

#### Proof Table:

| Statements | Reasons |
|-----------|--------|
| 1. $ \overline{BC} \cong \overline{DC} $, $ \overline{AC} \cong \overline{EC} $ | 1. Given |
| 2. $ \angle ACB \cong \angle ECD $ | 2. Vertical Angles Theorem |
| 3. $ \triangle BCA \cong \triangle DCE $ | 3. SAS |

✔ Answer for Problem 1:
- Postulate used: SAS
- Final proof completed.

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Problem 2

Given:
- $ \overline{JK} \cong \overline{LK} $
- $ \overline{JM} \cong \overline{LM} $

Prove: $ \triangle JKM \cong \triangle LKM $

#### Diagram Analysis:
- Draw triangle $ \triangle JKL $ with point $ M $ on base $ JL $.
- $ JK = LK $ → isosceles triangle?
- $ JM = LM $ → $ M $ is midpoint of $ JL $
- $ KM $ is common side to both triangles

Mark:
- $ JK \cong LK $ → one tick
- $ JM \cong LM $ → one tick
- $ KM $ is shared, so it’s congruent to itself → Reflexive Property

So we have:
- $ JK \cong LK $
- $ JM \cong LM $
- $ KM \cong KM $ (reflexive)

All three sides are congruent → SSS

#### Proof Table:

| 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 JKM \cong \triangle LKM $ | 3. SSS |

✔ Answer for Problem 2:
- Postulate used: SSS
- Final proof completed.

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Problem 3

Given:
- $ \angle G \cong \angle I $
- $ \overline{FH} $ bisects $ \angle GPI $

Prove: $ \triangle GFH \cong \triangle IFH $

#### Diagram Analysis:
- Triangle $ GPI $ with point $ H $ on $ GI $, and segment $ FH $ drawn from $ F $ to $ H $
- $ FH $ bisects $ \angle GPI $ → so $ \angle GFH \cong \angle IFH $
- $ \angle G \cong \angle I $ — given
- $ FH $ is common to both triangles → reflexive

We now look at the angles and sides:

From given:
- $ \angle G \cong \angle I $
- $ \angle GFH \cong \angle IFH $ (because $ FH $ bisects $ \angle GPI $)
- $ FH \cong FH $ (reflexive)

So we have:
- Two angles and the included side? Wait — actually, side $ FH $ is between $ \angle GFH $ and $ \angle GHF $? Not quite.

Wait — let's be careful.

We know:
- $ \angle G \cong \angle I $
- $ \angle GFH \cong \angle IFH $
- $ FH $ is common

So we have:
- $ \angle G $ and $ \angle GFH $ in $ \triangle GFH $
- $ \angle I $ and $ \angle IFH $ in $ \triangle IFH $
- Side $ FH $ is common

But this is AAS (Angle-Angle-Side) — two angles and a non-included side.

Alternatively, since $ FH $ is between the two angles in each triangle, wait — no:

In $ \triangle GFH $: angles at $ G $ and $ F $, side $ FH $
In $ \triangle IFH $: angles at $ I $ and $ F $, side $ FH $

So yes: two angles and the non-included side? Actually, side $ FH $ is not between the two angles unless we consider the vertex.

Wait — better to list:

- $ \angle G \cong \angle I $ → given
- $ \angle GFH \cong \angle IFH $ → definition of angle bisector
- $ FH \cong FH $ → reflexive

Now, these are two angles and a non-included side → AAS (Angle-Angle-Side)

Note: AAS is valid because if two angles are congruent, the third must be too (since sum is 180°), so it's equivalent to ASA.

But here, the side is not between the two angles — it's opposite one of them? No.

Actually, in $ \triangle GFH $, side $ FH $ is adjacent to $ \angle G $ and $ \angle GFH $. So it is included between $ \angle G $ and $ \angle GFH $? Wait — no.

Wait: $ \angle G $ is at vertex $ G $, $ \angle GFH $ is at $ F $. So side $ FH $ connects $ F $ and $ H $. So it is not between $ \angle G $ and $ \angle GFH $ — that would require the side to be $ GF $.

Actually, let's clarify:

- In $ \triangle GFH $:
- $ \angle G $ is at vertex $ G $
- $ \angle GFH $ is at vertex $ F $
- Side $ FH $ is opposite $ \angle G $

So $ FH $ is not between $ \angle G $ and $ \angle GFH $ — it's not adjacent to $ \angle G $?

Wait — $ \angle G $ has rays $ GF $ and $ GH $. So side $ GF $ and $ GH $ are adjacent to $ \angle G $.

But $ FH $ is from $ F $ to $ H $, so it’s not directly connected to $ G $ in the sense of being part of $ \angle G $.

So in $ \triangle GFH $, the side $ FH $ is opposite $ \angle G $.

Similarly, in $ \triangle IFH $, $ FH $ is opposite $ \angle I $.

So we have:
- $ \angle G \cong \angle I $
- $ \angle GFH \cong \angle IFH $
- $ FH \cong FH $

So: two angles and a non-included side → AAS

Yes! This is AAS.

Alternatively, you could argue ASA if you had the included side, but here $ FH $ is not the included side between the two angles.

But wait: $ \angle GFH $ and $ \angle IFH $ are adjacent to $ FH $, but $ \angle G $ and $ \angle I $ are not.

So the two angles are not both adjacent to $ FH $ — only one is.

Thus, AAS is correct.

Let’s complete the proof table.

#### Proof Table:

| Statements | Reasons |
|-----------|--------|
| 1. $ \angle G \cong \angle I $; $ \overline{FH} $ bisects $ \angle GPI $ | 1. Given |
| 2. $ \angle GFH \cong \angle IFH $ | 2. Definition of Angle Bisector |
| 3. $ \overline{FH} \cong \overline{FH} $ | 3. Reflexive Property |
| 4. $ \triangle GFH \cong \triangle IFH $ | 4. AAS |

✔ Answer for Problem 3:
- Postulate used: AAS
- Final proof completed.

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✔ Final Answers Summary:

#### Problem 1:
- Congruent parts: $ BC \cong DC $, $ AC \cong EC $, $ \angle ACB \cong \angle ECD $ (vertical angles)
- Postulate: SAS
- Proof:
1. Given
2. Vertical Angles Theorem
3. SAS

#### Problem 2:
- Congruent parts: $ JK \cong LK $, $ JM \cong LM $, $ KM \cong KM $
- Postulate: SSS
- Proof:
1. Given
2. Reflexive Property
3. SSS

#### Problem 3:
- Congruent parts: $ \angle G \cong \angle I $, $ \angle GFH \cong \angle IFH $, $ FH \cong FH $
- Postulate: AAS
- Proof:
1. Given
2. Definition of Angle Bisector
3. Reflexive Property
4. AAS

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Let me know if you'd like the diagrams annotated or need help drawing them!