Worksheet for practicing congruent triangle proofs with diagrams and structured proof tables.

It looks like you've uploaded a worksheet titled "Congruent Triangles: Challenge Proofs", which is designed to help students practice geometric proofs involving triangle congruence. However, the image itself does not contain a specific problem or diagram that can be solved directly — it's a blank template with spaces for statements and justifications in two separate proofs.

Let me explain what this worksheet is for and how to approach such problems, based on the structure visible in the image.

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🔍 Understanding the Worksheet

The worksheet has two columns of proof tables (each with 16 rows), labeled:

- Statement (left)
- Justification (right)

Each table corresponds to a different proof. There are two diagrams shown at the top:

#### 🔹 Diagram 1:
- Triangle $ \triangle PQR $
- Line segment $ TU $ is parallel to $ PQ $
- $ T $ is the midpoint of $ PR $
- $ U $ is the midpoint of $ QR $
- Goal: Prove $ \triangle PTU \cong \triangle QRU $? (Likely typo; probably meant $ \triangle PTU \cong \triangle QUR $ or similar.)

Wait — actually, looking more closely, the goal says:
> Prove: $ \triangle PTU \cong \triangle QRU $?
But from the diagram and information, it might be intended to prove $ \triangle PTU \cong \triangle QUR $ or perhaps $ \triangle PTU \cong \triangle QUT $? Let's clarify.

Actually, the given info says:
- $ TU \parallel PQ $
- $ T $ is midpoint of $ PR $
- $ U $ is midpoint of $ QR $

So we can use Midsegment Theorem or properties of parallel lines and midpoints.

But the proof goal is stated as:
> Prove: $ \triangle PTU \cong \triangle QRU $? → This seems incorrect unless there’s a typo.

Wait — let’s check again:
In the first diagram, points are:
- $ T $ on $ PR $
- $ U $ on $ QR $
- $ TU \parallel PQ $
- $ T $ and $ U $ are midpoints

So by the Midsegment Theorem, $ TU \parallel PQ $ and $ TU = \frac{1}{2}PQ $. Also, $ PT = TR $, $ QU = UR $

Now, consider triangles $ \triangle PTU $ and $ \triangle QUR $?

But the goal says: Prove: $ \triangle PTU \cong \triangle QRU $?

Let’s suppose the goal is to prove $ \triangle PTU \cong \triangle QUR $ — but even then, we need to analyze carefully.

Alternatively, maybe the goal is to prove something about segments, like $ PU = QT $? But the goal says “Prove: $ \triangle PTU \cong \triangle QRU $” — so likely a typo.

But wait — look at the second diagram.

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#### 🔹 Diagram 2:
- Triangle $ \triangle ABC $
- Point $ D $ is on $ AB $, point $ E $ is on $ AC $
- $ AD = DE $
- $ BD = EC $
- $ AE = DC $
- Goal: Prove: $ \triangle ABD \cong \triangle ACE $?

Wait — the goal says: Prove: $ \triangle ABD \cong \triangle ACE $?

But the given:
- $ AD = DE $
- $ BD = EC $
- $ AE = DC $

This is confusing because:
- $ AD = DE $ implies $ D $ is between $ A $ and $ E $? But $ E $ is on $ AC $
- $ BD = EC $
- $ AE = DC $

This suggests a symmetric construction.

Let’s interpret this properly.

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Since the image doesn't show actual content filled in, I'll walk through how to solve such a challenge proof using typical logic.

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✔ Step-by-Step Guide to Solving These Types of Problems

Let’s assume the first proof is:

> Given:
> - $ TU \parallel PQ $
> - $ T $ is midpoint of $ PR $
> - $ U $ is midpoint of $ QR $
> Prove: $ \triangle PTU \cong \triangle QUR $?
> Wait — better: perhaps prove $ \triangle PTU \cong \triangle QUR $? Or maybe $ \triangle PTU \cong \triangle RQU $? Not clear.

But actually, standard problem:
If $ T $ and $ U $ are midpoints, and $ TU \parallel PQ $, then $ TU $ is the midsegment of triangle $ \triangle PQR $, so:
- $ TU \parallel PQ $
- $ TU = \frac{1}{2}PQ $

And since $ T $ and $ U $ are midpoints:
- $ PT = TR $
- $ QU = UR $

Then consider $ \triangle PTU $ and $ \triangle RQU $? Still not matching.

Wait — perhaps the goal is to prove $ \triangle PTU \cong \triangle QUR $? That would require showing:
- $ PT = QU $? Not necessarily true
- $ TU = UR $? Only if triangle is isosceles

This is ambiguous.

But here’s a better interpretation:

Maybe the goal is to prove that $ \triangle PTU \cong \triangle QUR $, but that’s unlikely unless additional info is given.

Alternatively, perhaps the goal is to prove that $ \triangle PTU \cong \triangle QUR $ using SAS or ASA.

But without angles, it's hard.

Wait — let's try a different approach.

Perhaps the first problem is:

> Given:
> - $ TU \parallel PQ $
> - $ T $ is midpoint of $ PR $
> - $ U $ is midpoint of $ QR $
> Prove: $ \triangle PTU \cong \triangle QUR $?

No — better idea: maybe the goal is to prove $ \triangle PTU \cong \triangle QUR $ via SAS?

Let’s find equal sides and angles.

From midpoints:
- $ PT = TR $
- $ QU = UR $

But $ PT $ and $ QU $ are not necessarily equal.

Wait — perhaps the goal is to prove $ \triangle PTU \cong \triangle RQU $?

Still unclear.

But here’s a common problem:
Given $ T $ and $ U $ are midpoints, $ TU \parallel PQ $, then $ \triangle PTU \sim \triangle PQR $, but not necessarily congruent.

So perhaps the goal is not $ \triangle PTU \cong \triangle QRU $, but rather something else.

Wait — look again at the text:

> Prove: $ \triangle PTU \cong \triangle QRU $?

But $ \triangle QRU $ has points $ Q, R, U $. So vertices: $ Q, R, U $

Compare to $ \triangle PTU $: $ P, T, U $

So unless $ P = Q $, $ T = R $, etc., they’re not the same.

This suggests a typo in the goal.

A more likely goal is: Prove $ \triangle PTU \cong \triangle QUR $? But still, no clear correspondence.

Alternatively, maybe the goal is to prove $ \triangle PTU \cong \triangle RUT $? No.

Another possibility: maybe the goal is to prove $ \triangle PTU \cong \triangle QUR $, but only if we have symmetry.

Wait — perhaps the diagram shows that $ \triangle PQR $ is isosceles with $ PQ = PR $? But not stated.

Without more info, we can’t proceed.

Let’s move to the second proof.

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🔸 Second Proof

Given:
- $ AD = DE $
- $ BD = EC $
- $ AE = DC $
- Prove: $ \triangle ABD \cong \triangle ACE $

Let’s analyze.

We are to prove $ \triangle ABD \cong \triangle ACE $

Let’s list what we know:

1. $ AD = DE $ → So $ D $ lies on $ AE $, and $ AD = DE $, so $ D $ is midpoint of $ AE $
2. $ BD = EC $
3. $ AE = DC $

Wait — $ AE = DC $? That’s strange.

Let’s sketch mentally:

- Points: $ A, B, C $
- $ D $ on $ AB $, $ E $ on $ AC $
- $ AD = DE $
- $ BD = EC $
- $ AE = DC $

But $ AE $ is a segment from $ A $ to $ E $, and $ DC $ is from $ D $ to $ C $

So $ AE = DC $ means lengths are equal.

Now, we want to prove $ \triangle ABD \cong \triangle ACE $

Let’s see:

In $ \triangle ABD $ and $ \triangle ACE $:

- $ AB $ vs $ AC $? Not known
- $ AD $ vs $ AE $? $ AD = DE $, so $ AE = AD + DE = 2AD $, so $ AE = 2AD $
- $ BD = EC $ (given)
- $ AE = DC $ (given)

But $ DC $ is not a side of $ \triangle ACE $

Wait — perhaps we need to use triangle $ \triangle ABD $ and $ \triangle ACE $

Let’s try to find corresponding parts.

Suppose we want to prove $ \triangle ABD \cong \triangle ACE $

We need three pairs of congruent parts.

We know:
- $ BD = EC $ (given)
- $ AD = ? $ vs $ AE $? Not equal unless $ D $ is midpoint of $ AE $
- But $ AD = DE $, so $ AE = 2AD $

Also, $ AE = DC $ → so $ DC = AE = 2AD $

Now, let’s consider $ \triangle ABD $ and $ \triangle ACE $:

- $ BD = EC $ (given)
- $ AD = ? $ — not equal to $ AE $
- $ AB = ? $ — unknown
- $ AC = ? $ — unknown

But we also have $ AE = DC $

So $ AE = DC $

But $ AE $ is part of $ \triangle ACE $, $ DC $ is from $ D $ to $ C $

Now, consider $ \triangle ABD $ and $ \triangle ACE $:

Can we use SSS?

We need:
- $ AB = AC $? Not given
- $ BD = CE $? Yes, $ BD = EC $
- $ AD = AE $? No, $ AD = DE $, so $ AE = 2AD $

So $ AD \neq AE $

Thus SSS fails.

What about SAS?

We need two sides and included angle.

But we don’t have angles.

Wait — perhaps we need to consider other triangles.

Maybe the goal is to prove $ \triangle ABD \cong \triangle ACE $ using SAS?

Let’s suppose $ \angle BAD = \angle CAE $? Not given.

Alternatively, perhaps $ \triangle ABD \cong \triangle ACE $ via SSS if we can show $ AB = AC $, $ BD = CE $, $ AD = AE $? But $ AD \neq AE $

Unless $ D $ is not on $ AB $, but on $ AE $? Confusing.

Wait — the diagram shows:

- $ D $ on $ AB $
- $ E $ on $ AC $
- $ AD = DE $
- $ BD = EC $
- $ AE = DC $

Ah! So $ D $ is on $ AB $, $ E $ is on $ AC $

But $ AD = DE $ — so $ DE $ is a segment from $ D $ to $ E $, so $ AD = DE $ means length $ AD = $ length $ DE $

Similarly, $ BD = EC $, $ AE = DC $

Now, we want to prove $ \triangle ABD \cong \triangle ACE $

Let’s try to find three equal parts.

Let’s write down:

In $ \triangle ABD $ and $ \triangle ACE $:

1. $ BD = EC $ (given) → one pair of sides
2. $ AD = DE $ (given) → but $ DE $ is not a side of $ \triangle ACE $
3. $ AE = DC $ (given) → $ AE $ is side of $ \triangle ACE $, $ DC $ is not of $ \triangle ABD $

But $ DC $ is from $ D $ to $ C $, so it’s not a side of $ \triangle ABD $

So not helpful.

Wait — perhaps the goal is to prove $ \triangle ABD \cong \triangle ACE $ using SSS?

We need:
- $ AB = AC $? Not given
- $ BD = CE $? Yes
- $ AD = AE $? No

But $ AD = DE $, and $ AE = AD + DE = 2AD $, so $ AE = 2AD $, so $ AD \neq AE $

So $ AD \neq AE $

So not SSS.

Unless... maybe $ \triangle ABD \cong \triangle ACE $ is not correct.

Wait — perhaps the goal is $ \triangle ABD \cong \triangle ACE $, but only if we can show $ AB = AC $, $ BD = CE $, $ AD = AE $ — but $ AD \neq AE $

So unless there's a typo.

Wait — another idea: maybe the goal is $ \triangle ABD \cong \triangle ACE $, but using SAS with angles?

But no angles are given.

Perhaps we need to use the fact that $ AE = DC $, and $ BD = EC $, and $ AD = DE $

Let’s try to work with lengths.

Let $ AD = x $. Then $ DE = x $, so $ AE = AD + DE = 2x $

Given $ AE = DC $, so $ DC = 2x $

Also, $ BD = EC $, let’s say $ BD = y $, then $ EC = y $

Now, in $ \triangle ABD $: sides are $ AB = AD + DB = x + y $, $ AD = x $, $ BD = y $

In $ \triangle ACE $: $ AC = AE + EC = 2x + y $, $ AE = 2x $, $ EC = y $

So $ AB = x + y $, $ AC = 2x + y $ → not equal

So $ AB \neq AC $

So $ \triangle ABD $ and $ \triangle ACE $ are not congruent unless $ x = 0 $, impossible.

Therefore, the goal $ \triangle ABD \cong \triangle ACE $ is likely incorrect.

Perhaps the goal is to prove $ \triangle ABD \cong \triangle CED $? Or $ \triangle ADE \cong \triangle CED $? Not clear.

Alternatively, maybe the goal is to prove $ \triangle ABD \cong \triangle ACE $ using some transformation.

But without more info, it’s impossible.

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🚨 Conclusion

The worksheet appears to be a blank template for proving triangle congruence, but:

- The diagrams and goals are incomplete or possibly misprinted
- The statements are missing — only the setup is given
- Without specific values or a clear diagram, we cannot fill in the proof

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✔ How to Solve Such Problems (General Strategy)

Here’s how to approach these types of proofs:

#### 1. Identify Given Information
- List all given facts (e.g., midpoints, parallel lines, equal segments)
- Mark them on the diagram

#### 2. Use Definitions and Theorems
- Midpoint → divides segment into two equal parts
- Parallel lines → alternate interior angles equal
- Vertical angles are equal
- Reflexive property (e.g., $ AB = AB $)

#### 3. Look for Congruence Criteria
- SSS (Side-Side-Side)
- SAS (Side-Angle-Side)
- ASA (Angle-Side-Angle)
- AAS (Angle-Angle-Side)
- HL (for right triangles)

#### 4. Write the Proof in Two-Column Format
| Statement | Justification |
|---------|-------------|
| 1. $ T $ is midpoint of $ PR $ | Given |
| 2. $ PT = TR $ | Definition of midpoint |
| 3. $ TU \parallel PQ $ | Given |
| 4. $ \angle PTU = \angle PQB $ | Corresponding angles |
| ... | ... |

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🛠️ Example: First Problem (Assuming Corrected Goal)

Let’s assume the goal is to prove that $ \triangle PTU \cong \triangle QUR $, but only if $ \triangle PQR $ is isosceles.

But without that, it’s not possible.

Alternatively, a common problem is:

> Given: $ T $ and $ U $ are midpoints of $ PR $ and $ QR $, $ TU \parallel PQ $
> Prove: $ \triangle PTU \sim \triangle PQR $ (similar), not congruent

Or, prove $ TU = \frac{1}{2}PQ $ — which is the Midsegment Theorem.

So perhaps the goal is not congruence, but similarity or a segment equality.

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📌 Final Answer

Since the image contains a blank proof worksheet with no filled-in information or clear problem statement, I cannot solve a specific problem without additional details.

However, here’s how you should proceed:

1. Fill in the given information in the "Statement" column.
2. Use definitions, postulates, and theorems for justifications.
3. Work step by step toward the conclusion.
4. Use congruence criteria (SSS, SAS, etc.) when applicable.

If you can provide:
- The actual diagram
- The complete problem statement
- Or clarify the goal of the proof

Then I can give a full solution.

Would you like me to create an example proof for a typical "midpoint and parallel lines" problem?