Question

nick proves that any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment. his proof is shown. given: \\( \overline{j k} \\) is the perpendicular bisector of \\( \overline{l m} \\) prove: \\( \overline{l p} \cong \overline{m p} \\) 1. \\( \overline{j k} \\) is the perpendicular bisector of \\( \overline{l m} \\) 1. given 2. \\( \overline{k l} \cong \overline{k m} \\) 2. definition of perpendicular bisector 3. \\( \angle p k l \cong \angle p k m \\) 3. right angles are congruent 4. \\( \overline{k p} \cong \overline{k p} \\) 4. reflexive property of congruence 5. \\( \triangle k l p \cong \triangle k m p \\) 5. 6. \\( \overline{l p} \cong \overline{m p} \\) 6. corresponding parts of congruent triangles are congruent

Explanation:

Step1: Analyze the given information

We have \( \overline{KL}\cong\overline{KM}\) (from the definition of perpendicular bisector), \( \angle PKL\cong\angle PKM\) (right - angles are congruent), and \( \overline{KP}\cong\overline{KP}\) (reflexive property).

Step2: Apply the congruence criterion

Since we have two sides and the included angle (SAS - Side - Angle - Side) congruent. For \( \triangle KLP\) and \( \triangle KMP\), \( \overline{KL}\cong\overline{KM}\), \( \angle PKL\cong\angle PKM\), \( \overline{KP}\cong\overline{KP}\). So, by the SAS (Side - Angle - Side) congruence criterion, \( \triangle KLP\cong\triangle KMP\).

Answer:

1. SAS (Side - Angle - Side)