Question

proof point y is the midpoint of xz. point w is collinear with x, y, and z. z is the midpoint of yw. complete the two - column proof to prove that $overline{xy} cong overline{zw}$. drag the statements and reasons to complete the proof.

statements	reasons
2.	2. midpoint thm.
3.	3. def. of $cong$ segments
4.	4. transitive prop. of

Explanation:

Step 1: Identify Given Information

The given information is: \( Y \) is the midpoint of \( \overline{XZ} \), \( W \) is collinear with \( X, Y, Z \), and \( Z \) is the midpoint of \( \overline{YW} \). The reason for this statement is "Given".

Step 2: Apply Midpoint Theorem (for \( Y \) as midpoint of \( XZ \))

By the Midpoint Theorem, if \( Y \) is the midpoint of \( \overline{XZ} \), then \( XY = YZ \). So the statement for step 2 is \( XY = YZ \).

Step 3: Apply Midpoint Theorem (for \( Z \) as midpoint of \( YW \))

By the Midpoint Theorem, if \( Z \) is the midpoint of \( \overline{YW} \), then \( YZ = ZW \). So the statement for step 3 is \( YZ = ZW \).

Step 4: Apply Transitive Property of Equality

The Transitive Property of Equality states that if \( a = b \) and \( b = c \), then \( a = c \). Here, since \( XY = YZ \) (from step 2) and \( YZ = ZW \) (from step 3), we can conclude that \( XY = ZW \). By the definition of congruent segments (if two segments have equal length, they are congruent), \( \overline{XY} \cong \overline{ZW} \). Wait, actually, the transitive property here is for equality, and then we use the definition of congruent segments. But let's structure the two - column proof properly:

Filling the Two - Column Proof:

| Statements | Reasons |
| --- | --- |
| 1. \( Y \) is the midpoint of \( \overline{XZ} \). \( W \) is collinear with \( X, Y, Z \). \( Z \) is the midpoint of \( \overline{YW} \). | 1. Given |
| 2. \( XY = YZ \) | 2. Midpoint Thm. (If a point is the midpoint of a segment, it divides the segment into two equal - length segments) |
| 3. \( YZ = ZW \) | 3. Midpoint Thm. (Same reason as above, applied to \( Z \) as midpoint of \( YW \)) |
| 4. \( XY = ZW \) | 4. Transitive Prop. of Equality (If \( a = b \) and \( b = c \), then \( a = c \); here \( a = XY \), \( b = YZ \), \( c = ZW \)) |
| 5. \( \overline{XY} \cong \overline{ZW} \) | 5. Def. of \( \cong \) segments (If two segments have equal length, they are congruent) |

But since the problem has up to step 4 with reason "Transitive Prop. of", we can adjust:

After step 3 (\( YZ = ZW \)) and step 2 (\( XY = YZ \)), step 4: \( XY = ZW \) (Transitive Prop. of Equality), and then we can say \( \overline{XY} \cong \overline{ZW} \) by definition of congruent segments. But following the given table structure:

1. Reason for statement 1: Given
2. Statement 2: \( XY = YZ \) (Reason: Midpoint Thm.)
3. Statement 3: \( YZ = ZW \) (Reason: Midpoint Thm. - wait, no, the third reason is "Def. of \( \cong \) segments", so maybe:

Wait, the problem's table has:

- Row 1: Statement 1 (given), Reason 1 (to be filled as "Given")
- Row 2: Statement 2 (to be filled), Reason 2: Midpoint Thm.
- Row 3: Statement 3 (to be filled), Reason 3: Def. of \( \cong \) segments
- Row 4: Statement 4 (to be filled), Reason 4: Transitive Prop. of

Let's re - evaluate:

1. **Statement 1 Reason**: Given (because it's the information provided in the problem)
2. **Statement 2**: Since \( Y \) is the midpoint of \( \overline{XZ} \), by Midpoint Theorem, \( XY = YZ \). So Statement 2: \( XY = YZ \)
3. **Statement 3**: Since \( Z \) is the midpoint of \( \overline{YW} \), by Midpoint Theorem, \( YZ = ZW \). But the reason for row 3 is "Def. of \( \cong \) segments", which is about if two segments are congruent, their lengths are equal (or vice - versa). Wait, maybe there's a mix - up. Alternatively, after getting \( XY = YZ \) and \( YZ = ZW \), we can say \( XY = ZW \) (Transitive Prop.), and then \( \overline{XY} \cong \overline{ZW} \) (Def. of \( \cong \) segments).

Let's correct the two - column proof:

| Statements | Reasons |
| --- | --- |
| 1. \( Y \) is the midpoint of \( \overline{XZ} \); \( W \) is collinear with \( X, Y, Z \); \( Z \) is the midpoint of \( \overline{YW} \) | 1. Given |
| 2. \( XY = YZ \) | 2. Midpoint Theorem (A midpoint divides a segment into two congruent segments, so their lengths are equal) |
| 3. \( YZ = ZW \) | 3. Midpoint Theorem (Same as above, for \( Z \) as midpoint of \( YW \)) |
| 4. \( XY = ZW \) | 4. Transitive Property of Equality (If \( a = b \) and \( b = c \), then \( a = c \)) |
| 5. \( \overline{XY} \cong \overline{ZW} \) | 5. Definition of Congruent Segments (If two segments have equal length, they are congruent) |

But since the problem's table has up to row 4 with reason "Transitive Prop. of", we can assume that after showing \( XY = YZ \) and \( YZ = ZW \), we use transitive property to get \( XY = ZW \), and then by definition of congruent segments, \( \overline{XY} \cong \overline{ZW} \). But to fit the table:

1. Reason for row 1: Given
2. Statement for row 2: \( XY = YZ \) (Reason: Midpoint Thm.)
3. Statement for row 3: \( YZ = ZW \) (Reason: Midpoint Thm. - but the given reason for row 3 is "Def. of \( \cong \) segments", so maybe the problem has a different flow. Alternatively, maybe:

Wait, the "Def. of \( \cong \) segments" is that \( \overline{AB} \cong \overline{CD} \) iff \( AB = CD \). So maybe:

- Row 2: \( \overline{XY} \cong \overline{YZ} \) (Reason: Midpoint Thm. - since midpoint creates congruent segments)
- Row 3: \( \overline{YZ} \cong \overline{ZW} \) (Reason: Midpoint Thm. - \( Z \) is midpoint of \( YW \))
- Row 4: \( \overline{XY} \cong \overline{ZW} \) (Reason: Transitive Prop. of Congruence - if \( \overline{XY} \cong \overline{YZ} \) and \( \overline{YZ} \cong \overline{ZW} \), then \( \overline{XY} \cong \overline{ZW} \))

Yes, that makes more sense. Let's redo:

| Statements | Reasons |
| --- | --- |
| 1. \( Y \) is the midpoint of \( \overline{XZ} \); \( W \) is collinear with \( X, Y, Z \); \( Z \) is the midpoint of \( \overline{YW} \) | 1. Given |
| 2. \( \overline{XY} \cong \overline{YZ} \) | 2. Midpoint Theorem (A midpoint of a segment divides it into two congruent segments) |
| 3. \( \overline{YZ} \cong \overline{ZW} \) | 3. Midpoint Theorem ( \( Z \) is the midpoint of \( \overline{YW} \), so it divides \( \overline{YW} \) into two congruent segments) |
| 4. \( \overline{XY} \cong \overline{ZW} \) | 4. Transitive Property of Congruence (If \( \overline{AB} \cong \overline{BC} \) and \( \overline{BC} \cong \overline{CD} \), then \( \overline{AB} \cong \overline{CD} \)) |

Answer:

Filled Two - Column Proof:

Statements	Reasons
2. \( \overline{XY} \cong \overline{YZ} \)	2. Midpoint Thm.
3. \( \overline{YZ} \cong \overline{ZW} \)	3. Midpoint Thm. (Note: If the problem insists on "Def. of \( \cong \) segments" for row 3, there might be a misprint, but following the midpoint logic)
4. \( \overline{XY} \cong \overline{ZW} \)	4. Transitive Prop. of Congruence

(If we strictly follow the given reason for row 3 as "Def. of \( \cong \) segments", we can adjust row 2 to \( XY = YZ \) (Midpoint Thm.), row 3 to \( YZ = ZW \) (Midpoint Thm.), row 4 to \( XY = ZW \) (Transitive Prop. of Equality), and then a row 5 to \( \overline{XY} \cong \overline{ZW} \) (Def. of \( \cong \) segments). But based on the given table with 4 rows, the above is the best fit.)