Question

point m is the midpoint of \\(\overline{pq}\\), and \\(\overleftrightarrow{lm}\\) is the perpendicular bisector of \\(\overline{pq}\\). write a two - column proof to show that \\(lp = lq\\).
statements

1. \\(pm = qm\\), \\(lm\perp pq\\)
2. \\(lm = lm\\)
3. \\(\angle lpm = \angle lqm\\)
4. \\(\triangle pml\cong\triangle qml\\)
5. \\(\overline{lm}=\overline{lm}\\)
6. \\(lp = lq\\)

reasons

1. given
2. reflexive property of congruence
3. right angles congruence theorem
4. definition of segment congruence
5. corresponding parts of congruent triangles are congruent
6. hl congruence theorem

Explanation:

Step1: Analyze Statement 2

We know that \( \overline{LM} \) is a common side to both \( \triangle LPM \) and \( \triangle LQM \). By the Reflexive Property of Congruence, a segment is congruent to itself. So \( \overline{LM} \cong \overline{LM} \), which fills the second statement as \( \overline{LM}=\overline{LM} \) (using the definition of segment congruence for equality in length).

Step2: Analyze Statement 3

Since \( \overline{PN} \perp \overline{PQ} \) and \( \overline{LM} \) is the perpendicular bisector, \( \angle LMP \) and \( \angle LMQ \) are right angles. By the Right Angles Congruence Theorem, all right angles are congruent, so \( \angle LMP \cong \angle LMQ \).

Step3: Analyze Statement 4

We have \( \overline{PM}=\overline{QM} \) (given that \( M \) is the midpoint), \( \overline{LM}=\overline{LM} \) (reflexive), and \( \angle LMP \cong \angle LMQ \) (right angles). By the SAS (Side - Angle - Side) Congruence Theorem? Wait, no, wait. Wait, the triangles \( \triangle LPM \) and \( \triangle LQM \): \( \overline{PM}=\overline{QM} \), \( \angle LMP=\angle LMQ = 90^{\circ} \), \( \overline{LM}=\overline{LM} \). So by SAS, \( \triangle LPM\cong\triangle LQM \). Wait, but the reason for statement 4 is "Definition of segment congruence"? No, wait, the statement 4 is \( \triangle LPM\cong\triangle LQM \), and the reason should be SAS? Wait, maybe I misread. Wait, the given is \( \overline{PM}=\overline{QM} \), \( \overline{LM}\perp\overline{PQ} \), so \( \angle LMP \) and \( \angle LMQ \) are right angles (so \( \angle LMP\cong\angle LMQ \)), and \( \overline{LM} \) is common. So by SAS, \( \triangle LPM\cong\triangle LQM \). Then statement 4: \( \triangle LPM\cong\triangle LQM \), reason: SAS Congruence Theorem? But the given reason options? Wait, maybe the problem is about HL? Wait, no, \( \triangle LPM \) and \( \triangle LQM \) are right - angled at \( M \), \( \overline{PM}=\overline{QM} \), \( \overline{LM} \) is hypotenuse? Wait, no, \( \overline{LM} \) is a leg? Wait, maybe I made a mistake. Let's re - establish:

1. **Statement 1**: \( \overline{PM}=\overline{QM} \), \( \overline{LM}\perp\overline{PQ} \) (Given)
2. **Statement 2**: \( \overline{LM}=\overline{LM} \) (Reflexive Property of Congruence)
3. **Statement 3**: \( \angle LMP\cong\angle LMQ \) (Right Angles Congruence Theorem, since \( \overline{LM}\perp\overline{PQ} \), so both angles are \( 90^{\circ} \))
4. **Statement 4**: \( \triangle LPM\cong\triangle LQM \) (SAS Congruence Theorem: \( \overline{PM}=\overline{QM} \), \( \angle LMP=\angle LMQ \), \( \overline{LM}=\overline{LM} \))
5. **Statement 5**: \( \overline{LP}=\overline{LQ} \) (Corresponding Parts of Congruent Triangles are Congruent, CPCTC)
6. **Statement 6**: \( LP = LQ \) (Definition of Segment Congruence, since \( \overline{LP}\cong\overline{LQ} \) implies \( LP = LQ \))

Wait, but the table has:

- Statement 2: \( \overline{LM}=\overline{LM} \), Reason: Reflexive Property of Congruence (correct)
- Statement 3: \( \angle LMP\cong\angle LMQ \), Reason: Right Angles Congruence Theorem (correct)
- Statement 4: \( \triangle LPM\cong\triangle LQM \), Reason: SAS Congruence Theorem (since we have two sides and the included angle: \( \overline{PM}=\overline{QM} \), \( \angle LMP=\angle LMQ \), \( \overline{LM}=\overline{LM} \))
- Statement 5: \( \overline{LP}=\overline{LQ} \), Reason: CPCTC (Corresponding Parts of Congruent Triangles are Congruent)
- Statement 6: \( LP = LQ \), Reason: Definition of Segment Congruence (since congruent segments have equal length)

But let's fill the table as per the given structure:

| Statements | Reasons |
| --- | --- |
| 1. \( \overline{PM}=\overline{QM} \), \( \overline{LM}\perp\overline{PQ} \) | 1. Given |
| 2. \( \overline{LM}=\overline{LM} \) | 2. Reflexive Property of Congruence |
| 3. \( \angle LMP\cong\angle LMQ \) | 3. Right Angles Congruence Theorem |
| 4. \( \triangle LPM\cong\triangle LQM \) | 4. SAS Congruence Theorem (or HL? Wait, \( \triangle LPM \) and \( \triangle LQM \) are right - angled, \( \overline{PM}=\overline{QM} \), \( \overline{LM} \) is common. So by HL, since \( \overline{LM} \) is the hypotenuse? Wait, \( \overline{LP} \) and \( \overline{LQ} \) are hypotenuses? Wait, no, \( \angle LMP \) and \( \angle LMQ \) are right angles, so the legs are \( \overline{PM},\overline{LM} \) and \( \overline{QM},\overline{LM} \), and hypotenuses are \( \overline{LP} \) and \( \overline{LQ} \). So if we consider \( \overline{PM}=\overline{QM} \), \( \overline{LM}=\overline{LM} \), then by HL (Hypotenuse - Leg) Congruence Theorem? Wait, HL is for right - angled triangles, where the hypotenuse and one leg of one triangle are congruent to the hypotenuse and one leg of another triangle. Here, \( \overline{LM} \) is a leg, not the hypotenuse. So SAS is correct. |
| 5. \( \overline{LP}=\overline{LQ} \) | 5. CPCTC (Corresponding Parts of Congruent Triangles are Congruent) |
| 6. \( LP = LQ \) | 6. Definition of Segment Congruence |

But the problem's table has:

- Statement 2: \( \overline{LM}=\overline{LM} \), Reason: Reflexive Property of Congruence (correct)
- Statement 3: \( \angle LMP\cong\angle LMQ \), Reason: Right Angles Congruence Theorem (correct)
- Statement 4: \( \triangle LPM\cong\triangle LQM \), Reason: Let's see, the given reason for statement 4 is "Definition of segment congruence"? No, that's not. Wait, maybe the original problem has a typo, but following the steps:

To show \( LP = LQ \), we prove \( \triangle LPM\cong\triangle LQM \) and then use CPCTC.

1. **Step 1 (Statement 2)**: Since \( \overline{LM} \) is a common segment to both \( \triangle LPM \) and \( \triangle LQM \), by the reflexive property of congruence, \( \overline{LM}\cong\overline{LM} \), so \( \overline{LM}=\overline{LM} \) (in terms of length, by definition of segment congruence).
2. **Step 2 (Statement 3)**: Because \( \overline{LM}\perp\overline{PQ} \), \( \angle LMP \) and \( \angle LMQ \) are right angles. By the right angles congruence theorem, all right angles are congruent, so \( \angle LMP\cong\angle LMQ \).
3. **Step 3 (Statement 4)**: We have \( \overline{PM}=\overline{QM} \) (given, \( M \) is the midpoint), \( \angle LMP\cong\angle LMQ \) (right angles), and \( \overline{LM}=\overline{LM} \) (reflexive). By the SAS (Side - Angle - Side) congruence theorem, \( \triangle LPM\cong\triangle LQM \).
4. **Step 4 (Statement 5)**: By the corresponding parts of congruent triangles are congruent (CPCTC) theorem, since \( \triangle LPM\cong\triangle LQM \), their corresponding sides \( \overline{LP} \) and \( \overline{LQ} \) are congruent, so \( \overline{LP}\cong\overline{LQ} \).
5. **Step 5 (Statement 6)**: By the definition of segment congruence, if two segments are congruent, their lengths are equal. So \( LP = LQ \).

Answer:

The completed two - column proof is as follows:

Statements	Reasons
2. \( \overline{LM}=\overline{LM} \)	2. Reflexive Property of Congruence
3. \( \angle LMP\cong\angle LMQ \)	3. Right Angles Congruence Theorem
4. \( \triangle LPM\cong\triangle LQM \)	4. SAS Congruence Theorem
5. \( \overline{LP}=\overline{LQ} \)	5. CPCTC
6. \( LP = LQ \)	6. Definition of Segment Congruence

(If we follow the given table structure in the problem, filling the blanks:

• Statement 2: \( \overline{LM}=\overline{LM} \)
• Statement 3: \( \angle LMP\cong\angle LMQ \)
• Statement 4: \( \triangle LPM\cong\triangle LQM \)
• Statement 5: \( \overline{LP}=\overline{LQ} \)
• Statement 6: \( LP = LQ \) with the respective reasons as above)