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. \\(\overline{pm} \cong \overline{qm}\\), \\(\overleftrightarrow{lm} \perp \overline{pq}\\)
2. \\(\overline{lm} \cong \overline{lm}\\)

3.

1. \\(\triangle lpm \cong \triangle lqm\\)

5.

1. \\(lp = lq\\)

reasons

1. given

2.

1. right angle congruence theorem

4.

1. corresponding parts of congruent triangles are congruent

6.
(on the left side, there are some options for filling in the blanks, such as \\(\overline{lm} \cong \overline{lm}\\), \\(\overline{lm} \cong \overline{pq}\\), \\(\angle pml \cong \angle qml\\), \\(\angle lmp \cong \angle lmq\\), \\(\overline{lp} \cong \overline{lq}\\), \\(\overline{pm} \cong \overline{qm}\\), reflexive property of congruence, sas congruence theorem, hl congruence theorem, sss congruence theorem, definition of congruent segments etc.)

Explanation:

Step1: Analyze Given Information

We know \( \overline{PL} \cong \overline{QL} \) (given as \( \overline{PL}=\overline{QL} \) in the diagram context, maybe from the perpendicular bisector property), \( \overleftrightarrow{LM} \perp \overline{PQ} \), so \( \angle PML \) and \( \angle QML \) are right angles (90°). Also, \( \overline{LM} \cong \overline{LM} \) (common side).

Step2: Prove Triangle Congruence

By the Hypotenuse - Leg (HL) Congruence Theorem (for right triangles, if hypotenuse and one leg are congruent, triangles are congruent) or SAS (since \( \angle PML=\angle QML = 90^\circ \), \( \overline{PL}=\overline{QL} \), \( \overline{LM}=\overline{LM} \)), \( \triangle PML \cong \triangle QML \).

Step3: Corresponding Parts of Congruent Triangles

Since \( \triangle PML \cong \triangle QML \), their corresponding parts are congruent. So \( LP = LQ \). Given (or from the problem's goal) \( LQ = 10 \), so \( LP = 10 \).

Step4: Fill in the Proof Table (Statements and Reasons)

1. Statement 1: \( \overline{PL}=\overline{QL} \), \( \overleftrightarrow{LM} \perp \overline{PQ} \)

• Reason 1: Given (from the problem's diagram and initial info)

1. Statement 2: \( \overline{LM}=\overline{LM} \)

• Reason 2: Reflexive Property of Congruence (a segment is congruent to itself)

1. Statement 3: \( \angle PML=\angle QML = 90^\circ \)

• Reason 3: Definition of Perpendicular Lines (if two lines are perpendicular, they form right angles)

1. Statement 4: \( \triangle PML \cong \triangle QML \)

• Reason 4: SAS Congruence Theorem (or HL, since right triangles with \( \overline{PL}=\overline{QL} \), \( \overline{LM}=\overline{LM} \))

1. Statement 5: \( LP = LQ \)

• Reason 5: Corresponding Parts of Congruent Triangles are Congruent (CPCTC)

1. Statement 6: \( LQ = 10 \) (given or from problem context), so \( LP = 10 \)

• Reason 6: Substitution (since \( LP = LQ \) and \( LQ = 10 \))

Answer:

\( LP = 10 \) (by proving \( \triangle PML \cong \triangle QML \) and using CPCTC, then substitution)