Question

unit 4a: the midpoint of \\(\overline{nq}\\), and \\(\overline{lm}\\) is the perpendicular bisector of \\(\overline{nq}\\). what is the last reason to prove that \\(lp = lq\\).\
statements\

1. \\(mo = mq\\), \\(lm \perp nq\\)\
2. \\(\overline{lm} \cong \overline{lm}\\)\
3. \
4. \\(\triangle lmn \cong \triangle lmq\\)\
5. \
6. \\(lp = lq\\)\

reasons\

1. given\
2. \
3. right angle congruence theorem\
4. \
5. corresponding parts of congruent triangles are congruent.\
6. \

there are also some congruence and equality statements on the right side of the image, like \\(\overline{lm} \cong \overline{lm}\\), \\(\triangle lmn \cong \triangle lmq\\), etc., and some other congruence theorems listed vertically.

Explanation:

Step1: Identify Given Information

We know \( \overline{IM} \) is the perpendicular bisector of \( \overline{NQ} \), so \( MN = MQ \) (perpendicular bisector definition: any point on the perpendicular bisector is equidistant from the endpoints of the segment), and \( \angle IMN=\angle IMQ = 90^\circ \) (perpendicular lines form right angles). Also, \( IM = IM \) (common side, reflexive property).

Step2: Apply Congruence Criterion

For triangles \( \triangle IMN \) and \( \triangle IMQ \), we have:

• \( MN = MQ \) (from perpendicular bisector)
• \( \angle IMN=\angle IMQ = 90^\circ \) (right angles)
• \( IM = IM \) (common side)

This satisfies the SAS (Side - Angle - Side) congruence theorem for right triangles (also known as the Hypotenuse - Leg theorem is for right triangles, but here we have two sides and the included angle, so SAS). So \( \triangle IMN\cong\triangle IMQ \) by SAS (or Right - Angle Congruence if considering right triangles).

Step3: Corresponding Parts of Congruent Triangles

Since \( \triangle IMN\cong\triangle IMQ \), their corresponding parts (angles) are congruent. The angles \( \angle IP \) (assuming \( \angle IP \) refers to \( \angle INP \) and \( \angle IQP \), but from the problem statement \( \angle LP=\angle LQ \), likely a typo and should be \( \angle NP=\angle QP \) or \( \angle INP=\angle IQP \)) are corresponding angles of congruent triangles. So by the Corresponding Parts of Congruent Triangles are Congruent (CPCTC) theorem, \( \angle NP=\angle QP \) (or \( \angle LP=\angle LQ \) as per the problem) because they are corresponding parts of \( \triangle IMN \) and \( \triangle IMQ \).

Answer:

The two - column proof can be filled as follows:

STATEMENTS	REASONS
2. \( MN = MQ \), \( \angle IMN=\angle IMQ = 90^\circ \)	2. Definition of perpendicular bisector (perpendicular bisector of a segment is a line that is perpendicular to the segment at its midpoint, so it creates two right angles and the midpoint divides the segment into two equal parts)
3. \( IM = IM \)	3. Reflexive Property of Equality
4. \( \triangle IMN\cong\triangle IMQ \)	4. SAS (Side - Angle - Side) Congruence Theorem (or Right - Angle Congruence Theorem for right triangles)
5. \( \angle NP=\angle QP \) (or \( \angle LP=\angle LQ \))	5. Corresponding Parts of Congruent Triangles are Congruent (CPCTC)

(If we assume the final answer is the congruence of angles \( \angle LP=\angle LQ \) due to CPCTC after proving \( \triangle IMN\cong\triangle IMQ \), then the key steps lead to the conclusion that \( \angle LP=\angle LQ \) because of the congruence of the triangles and CPCTC.)