Worksheet for proving triangle congruence with diagrams and proof tables.

Let's solve each problem step-by-step, following the instructions:

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Problem 1

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

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

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#### a. Show given information in diagram (tick marks):
- Mark $ \overline{BC} \cong \overline{DC} $ with one tick mark.
- Mark $ \overline{AC} \cong \overline{EC} $ with another tick mark.

#### b. Other congruent parts:
- $ \angle ACB \cong \angle ECD $ — these are vertical angles, so they are congruent by the Vertical Angles Theorem.

#### c. Congruence Postulate:
We have:
- $ \overline{BC} \cong \overline{DC} $
- $ \angle ACB \cong \angle ECD $
- $ \overline{AC} \cong \overline{EC} $

This is SAS (Side-Angle-Side) congruence.

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#### 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 ∠s Theorem |
| 3. $ \triangle BCA \cong \triangle DCE $ | 3. SAS |

✔ Answer: Use SAS postulate.

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Problem 2

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

Prove: $ \triangle KJM \cong \triangle KLM $

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#### a. Diagram markings:
- Mark $ \overline{JK} \cong \overline{LK} $ with one tick.
- Mark $ \overline{JM} \cong \overline{LM} $ with another tick.

#### b. Other congruent parts:
- $ \overline{KM} $ is shared by both triangles → $ \overline{KM} \cong \overline{KM} $ by Reflexive Property.

#### c. Congruence Postulate:
We have:
- $ \overline{JK} \cong \overline{LK} $
- $ \overline{JM} \cong \overline{LM} $
- $ \overline{KM} \cong \overline{KM} $

All three sides are congruent → SSS (Side-Side-Side)

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#### 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 Prop. |
| 3. $ \triangle KJM \cong \triangle KLM $ | 3. SSS |

✔ Answer: Use SSS postulate.

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Problem 3

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

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

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#### a. Diagram markings:
- Mark $ \angle G \cong \angle I $ with arcs.
- Since $ \overline{FH} $ bisects $ \angle GFI $, mark $ \angle GFH \cong \angle IFH $ with equal arcs.

#### b. Other congruent parts:
- $ \overline{FH} $ is shared → $ \overline{FH} \cong \overline{FH} $ by Reflexive Property.

#### c. Congruence Postulate:
We have:
- $ \angle G \cong \angle I $ (given)
- $ \angle GFH \cong \angle IFH $ (by angle bisector)
- $ \overline{FH} \cong \overline{FH} $ (reflexive)

So we have AAS (Angle-Angle-Side) — two angles and a non-included side.

Note: We could also use ASA if we had included side, but here $ FH $ is not between the two angles in both triangles? Let’s check:

In $ \triangle GFH $: angles at $ G $ and $ F $, side $ FH $ is opposite to $ G $? Wait — better to see:

Actually:
- $ \angle G \cong \angle I $
- $ \angle GFH \cong \angle IFH $
- $ \overline{FH} \cong \overline{FH} $

But $ \overline{FH} $ is between $ \angle GFH $ and $ \angle H $ in both triangles? Not quite.

Wait — let's label carefully.

In $ \triangle GFH $: angles at $ G $, $ F $, $ H $.
In $ \triangle IFH $: angles at $ I $, $ F $, $ H $.

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

So we have two angles and a non-included side (since $ FH $ is not between $ \angle G $ and $ \angle GFH $).

Wait — actually, $ \angle G $ and $ \angle GFH $ are adjacent, and $ FH $ is not the included side — it's opposite $ \angle G $? No.

Better: In triangle $ GFH $, the side $ FH $ is adjacent to $ \angle GFH $ and $ \angle H $, but not to $ \angle G $.

So actually, we have:
- $ \angle G \cong \angle I $
- $ \angle GFH \cong \angle IFH $
- $ \overline{FH} \cong \overline{FH} $

This is AAS: Two angles and a non-included side.

Yes — AAS applies.

Alternatively, since we have two angles and the included side?

Wait — no. The side $ FH $ is not between $ \angle G $ and $ \angle GFH $. It's between $ \angle GFH $ and $ \angle H $. So it's not the included side for $ \angle G $ and $ \angle GFH $.

So yes, AAS.

But wait — could we use ASA?

No — unless we had the included side.

But here, the side $ FH $ is not between the two known angles in either triangle.

Wait — let's re-evaluate.

In $ \triangle GFH $:
- $ \angle G $
- $ \angle GFH $
- $ \overline{FH} $

Is $ \overline{FH} $ adjacent to $ \angle G $? No — $ \angle G $ is at vertex $ G $, and $ FH $ connects $ F $ and $ H $, so it's not connected to $ G $ directly.

So $ \overline{FH} $ is opposite $ \angle G $? No — $ \angle G $ is at $ G $, so opposite side is $ FH $? Yes!

Wait — in triangle $ GFH $, side opposite $ \angle G $ is $ \overline{FH} $.

Similarly, in $ \triangle IFH $, side opposite $ \angle I $ is $ \overline{FH} $.

So we have:
- $ \angle G \cong \angle I $
- $ \angle GFH \cong \angle IFH $
- $ \overline{FH} \cong \overline{FH} $ — this is the side opposite $ \angle G $ and $ \angle I $

So this is AAS: two angles and a non-included side (the side opposite one of the angles).

Yes — AAS is valid.

Alternatively, since we have two angles, the third angle must be congruent, and then we can use ASA.

But standard is AAS.

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#### Proof Table:

| 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 |

✔ Answer: Use AAS postulate.

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✔ Final Answers Summary:

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

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#### Problem 2:
- Statements:
1. $ \overline{JK} \cong \overline{LK}, \overline{JM} \cong \overline{LM} $
2. $ \overline{KM} \cong \overline{KM} $
3. $ \triangle KJM \cong \triangle KLM $
- Reasons:
1. Given
2. Reflexive Prop.
3. SSS

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

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Let me know if you'd like the diagrams annotated with tick marks and arcs!