Student's completed segment proofs homework with handwritten answers and diagrams.

Let's go through the homework step by step and correctly complete the missing parts of the proofs, especially focusing on problems 4 and 5. I'll also verify the segment addition postulate equations in part 1–3.

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✔ Part 1–3: Segment Addition Postulate Equations

Using the diagram:

```
P ---- Q ---- R ---- S ---- T
```

We use the Segment Addition Postulate: If point $ B $ is between points $ A $ and $ C $, then $ AB + BC = AC $.

So the three correct equations are:

1. $ \overline{PR} + \overline{RT} = \overline{PT} $
→ This is correct (R is between P and T)

2. $ \overline{PQ} + \overline{QR} = \overline{PR} $
→ Correct (Q is between P and R)

3. $ \overline{PQ} + \overline{QS} + \overline{ST} = \overline{PT} $
→ Also correct (adding segments from P to T via Q, S)

✔ These are all valid applications of the segment addition postulate.

---

🔷 Problem 4: Proof

Given:
- $ X $ is the midpoint of $ \overline{WY} $
- $ \overline{WX} \cong \overline{XZ} $

Prove: $ \overline{XY} \cong \overline{XZ} $

#### Table:

| Statements | Reasons |
|-----------|--------|
| 1. $ X $ is the midpoint of $ \overline{WY} $ | 1. Given |
| 2. $ WX = XY $ | 2. Definition of Midpoint |
| 3. $ \overline{WX} \cong \overline{XZ} $ | 3. Given |
| 4. $ WX = XZ $ | 4. Definition of Congruence |
| 5. $ XY = XZ $ | 5. Transitive Property of Equality |
| 6. $ \overline{XY} \cong \overline{XZ} $ | 6. Definition of Congruence |

> ✔ Explanation:
- Step 2: Since $ X $ is the midpoint, $ WX = XY $ by definition.
- Step 3: Given that $ \overline{WX} \cong \overline{XZ} $
- Step 4: Convert congruence to equality: $ WX = XZ $
- Step 5: Use transitive property: $ XY = WX $ and $ WX = XZ $ ⇒ $ XY = XZ $
- Step 6: Convert equality back to congruence.

✔️ Final Answer: $ \overline{XY} \cong \overline{XZ} $ — Proven.

---

🔷 Problem 5: Proof

Given: $ \overline{AB} \cong \overline{CD} $

Prove: $ \overline{AC} \cong \overline{BD} $

Diagram:
```
A ---- B ---- C ---- D
```

So we have a line with points in order: A, B, C, D.

We need to prove $ \overline{AC} \cong \overline{BD} $

#### Table:

| Statements | Reasons |
|-----------|--------|
| 1. $ \overline{AB} \cong \overline{CD} $ | 1. Given |
| 2. $ AB = CD $ | 2. Definition of Congruence |
| 3. $ AC = AB + BC $ | 3. Segment Addition Postulate |
| 4. $ BD = BC + CD $ | 4. Segment Addition Postulate |
| 5. $ CD + BC = BD $ | 5. Commutative Property of Addition |
| 6. $ AC = AB + BC $ and $ BD = CD + BC $ | 6. Substitution (from steps 3 & 4) |
| 7. $ AC = CD + BC $ | 7. Substitution (since $ AB = CD $ from step 2) |
| 8. $ AC = BD $ | 8. Transitive Property of Equality |
| 9. $ \overline{AC} \cong \overline{BD} $ | 9. Definition of Congruence |

But the table given in the image has some different steps. Let’s match it exactly as written.

The provided table looks like this:

| Statements | Reasons |
|-----------|--------|
| 1. $ \overline{AB} \cong \overline{CD} $ | 1. Given |
| 2. $ AB = CD $ | 2. Definition of Congruence |
| 3. $ AC + CD = AD $ | 3. Segment Addition Postulate |
| 4. $ AB + BD = AD $ | 4. ??? |
| 5. $ CD + BD = AD $ | 5. ??? |
| 6. $ AC + CD = CD + BD $ | 6. ??? |
| 7. $ AC = BD $ | 7. Transitive Property |
| 8. $ \overline{AC} \cong \overline{BD} $ | 8. Definition of Congruence |

Wait — there's an issue here. The logic seems flawed because:

- Statement 4: $ AB + BD = AD $? That would only be true if B is between A and D and D is beyond B — but we don't know that. Actually, the diagram shows A-B-C-D, so B is not between A and D, but rather C is between B and D.

So let's fix this proof using proper reasoning.

---

✔ Corrected Version of Problem 5

We want to prove $ \overline{AC} \cong \overline{BD} $

Assume: Points are collinear in order: A – B – C – D

So:
- $ AC = AB + BC $
- $ BD = BC + CD $

Given: $ AB = CD $

So:
- $ AC = AB + BC $
- $ BD = CD + BC = AB + BC $ (since $ AB = CD $)
⇒ $ AC = BD $

Thus, $ \overline{AC} \cong \overline{BD} $

Now fill in the table correctly:

| Statements | Reasons |
|-----------|--------|
| 1. $ \overline{AB} \cong \overline{CD} $ | 1. Given |
| 2. $ AB = CD $ | 2. Definition of Congruence |
| 3. $ AC = AB + BC $ | 3. Segment Addition Postulate |
| 4. $ BD = BC + CD $ | 4. Segment Addition Postulate |
| 5. $ BD = BC + AB $ | 5. Substitution (from step 2) |
| 6. $ AC = AB + BC $ | 6. Step 3 |
| 7. $ AC = BD $ | 7. Transitive Property of Equality |
| 8. $ \overline{AC} \cong \overline{BD} $ | 8. Definition of Congruence |

But the original table uses a different approach involving $ AD $. Let’s analyze what they might have intended.

They wrote:

- 3. $ AC + CD = AD $ → OK (C between A and D)
- 4. $ AB + BD = AD $ → ✘ Incorrect! B is not between A and D? Wait — actually, if A-B-C-D, then B is between A and C, but not necessarily between A and D unless we assume something.

But if you consider that $ AB + BC + CD = AD $, then $ AB + BD = AD $? Only if $ BD = BC + CD $, which is true.

But $ AB + BD = AB + (BC + CD) = AB + BC + CD = AD $ → Yes!

So:

- $ AB + BD = AD $ is true if $ BD = BC + CD $

So let's rework it as per their structure:

| Statements | Reasons |
|-----------|--------|
| 1. $ \overline{AB} \cong \overline{CD} $ | 1. Given |
| 2. $ AB = CD $ | 2. Definition of Congruence |
| 3. $ AC + CD = AD $ | 3. Segment Addition Postulate (C between A and D) |
| 4. $ AB + BD = AD $ | 4. Segment Addition Postulate (B between A and D)? Wait — no! B is not between A and D unless we define it.

Actually, B is between A and C, not necessarily between A and D.

But D is beyond C, so B is not between A and D — so $ AB + BD = AD $ is invalid.

So this step is incorrect.

Therefore, the original table in the image has a logical flaw.

But perhaps the intention was:

- $ AC + CD = AD $ → OK
- $ AB + BC + CD = AD $ → OK
- But $ AB + BD $ isn’t valid.

Alternative idea: Maybe they meant $ AB + BC + CD = AD $, and $ AC = AB + BC $, $ BD = BC + CD $, etc.

But since the table shows:

| 4. $ AB + BD = AD $ | 4. ??? |

That’s not valid unless D is after B, but even then, $ AB + BD = AD $ only if B is between A and D — which is true only if D is after B, which it is — but wait: Is B between A and D?

Yes! In the order A-B-C-D, B is between A and D.

So yes, $ AB + BD = AD $ is valid by segment addition postulate, because B is between A and D.

Similarly, $ AC + CD = AD $, since C is between A and D.

So both are valid.

Let’s proceed carefully.

So:

- $ AB + BD = AD $ → B is between A and D → OK
- $ AC + CD = AD $ → C is between A and D → OK

So now:

From:
- $ AB + BD = AD $
- $ AC + CD = AD $

So: $ AB + BD = AC + CD $

But $ AB = CD $, so substitute:

→ $ CD + BD = AC + CD $

Subtract CD from both sides: $ BD = AC $

So $ AC = BD $, hence $ \overline{AC} \cong \overline{BD} $

Now fill in the table properly:

| Statements | Reasons |
|-----------|--------|
| 1. $ \overline{AB} \cong \overline{CD} $ | 1. Given |
| 2. $ AB = CD $ | 2. Definition of Congruence |
| 3. $ AC + CD = AD $ | 3. Segment Addition Postulate (C between A and D) |
| 4. $ AB + BD = AD $ | 4. Segment Addition Postulate (B between A and D) |
| 5. $ AC + CD = AB + BD $ | 5. Transitive Property of Equality (both equal AD) |
| 6. $ AC + CD = CD + BD $ | 6. Substitution (replace AB with CD) |
| 7. $ AC = BD $ | 7. Subtraction Property of Equality (subtract CD from both sides) |
| 8. $ \overline{AC} \cong \overline{BD} $ | 8. Definition of Congruence |

This matches the structure in the image.

Now, check the image’s version:

It says:

- 3. $ AC + CD = AD $ → ✔️
- 4. $ AB + BD = AD $ → ✔️ (if B is between A and D — which it is)
- 5. $ CD + BD = AD $ → ✘ This is not correct. It should be $ AB + BD = AD $, not $ CD + BD $

Wait — in the image, statement 5 is: $ CD + BD = AD $

But $ CD + BD $ is not equal to $ AD $ unless C is between A and D and B is between C and D — but B is before C.

So $ CD + BD $ is not a valid expression for $ AD $

So this is wrong.

Let’s compare:

Image says:

| 5. $ CD + BD = AD $ | 5. ??? |

But $ CD + BD $ is not $ AD $

For example: If AB=2, BC=3, CD=2, then AD = 2+3+2=7

Then:
- $ AB + BD = 2 + (3+2) = 7 $
- $ AC + CD = (2+3) + 2 = 7 $
- $ CD + BD = 2 + 5 = 7 $ → Wait! That’s also 7?

Is $ CD + BD = AD $? Let’s see:

- $ CD = 2 $
- $ BD = BC + CD = 3 + 2 = 5 $
- $ CD + BD = 2 + 5 = 7 $
- $ AD = 7 $

So numerically it works — but is this a valid application of segment addition?

No — because D is not between C and B, so $ CD + BD $ doesn’t represent a single segment.

In fact, $ CD + BD $ is adding two segments that overlap at D, but not forming a straight path from C to B.

So it’s not valid geometrically.

So the step $ CD + BD = AD $ is incorrect reasoning.

Therefore, the image has a mistake.

But let’s look again at the actual flow:

They write:

- 3. $ AC + CD = AD $ → OK
- 4. $ AB + BD = AD $ → OK (since B is between A and D)
- 5. $ CD + BD = AD $ → ✘ Not valid
- 6. $ AC + CD = CD + BD $ → OK if both equal AD
- 7. $ AC = BD $ → OK (subtract CD)
- 8. $ \overline{AC} \cong \overline{BD} $ → OK

But step 5 is problematic.

Instead, step 5 should be:

> $ AC + CD = AB + BD $ → From steps 3 and 4

Then step 6: $ AC + CD = CD + BD $ → Replace AB with CD (from step 2)

So:

| Statements | Reasons |
|-----------|--------|
| 1. $ \overline{AB} \cong \overline{CD} $ | 1. Given |
| 2. $ AB = CD $ | 2. Definition of Congruence |
| 3. $ AC + CD = AD $ | 3. Segment Addition Postulate |
| 4. $ AB + BD = AD $ | 4. Segment Addition Postulate |
| 5. $ AC + CD = AB + BD $ | 5. Transitive Property (both equal AD) |
| 6. $ AC + CD = CD + BD $ | 6. Substitution (replace AB with CD) |
| 7. $ AC = BD $ | 7. Subtraction Property of Equality |
| 8. $ \overline{AC} \cong \overline{BD} $ | 8. Definition of Congruence |

✔ This is correct.

So in the image, step 5 should be: $ AC + CD = AB + BD $, not $ CD + BD = AD $

But it says $ CD + BD = AD $ — which is numerically true but geometrically invalid.

So likely a typo or mislabeling.

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✔ Final Answers:

#### Problem 4: Complete

| Statements | Reasons |
|-----------|--------|
| 1. $ X $ is the midpoint of $ \overline{WY} $ | 1. Given |
| 2. $ WX = XY $ | 2. Definition of Midpoint |
| 3. $ \overline{WX} \cong \overline{XZ} $ | 3. Given |
| 4. $ WX = XZ $ | 4. Definition of Congruence |
| 5. $ XY = XZ $ | 5. Transitive Property of Equality |
| 6. $ \overline{XY} \cong \overline{XZ} $ | 6. Definition of Congruence |

#### Problem 5: Corrected

| Statements | Reasons |
|-----------|--------|
| 1. $ \overline{AB} \cong \overline{CD} $ | 1. Given |
| 2. $ AB = CD $ | 2. Definition of Congruence |
| 3. $ AC + CD = AD $ | 3. Segment Addition Postulate |
| 4. $ AB + BD = AD $ | 4. Segment Addition Postulate |
| 5. $ AC + CD = AB + BD $ | 5. Transitive Property of Equality |
| 6. $ AC + CD = CD + BD $ | 6. Substitution (AB = CD) |
| 7. $ AC = BD $ | 7. Subtraction Property of Equality |
| 8. $ \overline{AC} \cong \overline{BD} $ | 8. Definition of Congruence |

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✔ Summary

- Problems 1–3: All correct.
- Problem 4: Completed correctly.
- Problem 5: Original had error in step 5. Corrected version above.

Let me know if you'd like this typed neatly for submission!