Question

10
if \\( \angle l \cong \angle n \\), what else is needed to prove \\( \triangle k l m \cong \triangle p n m \\) using the angle-side-angle rule?
\\( \bigcirc \mathrm { a } \\). \\( \overline { \mathrm { km } } \cong \overline { \mathrm { pm } } \\)
\\( \bigcirc \mathrm { b } \\). \\( \overline { \mathrm { lm } } \cong \overline { \mathrm { nm } } \\)
\\( \bigcirc \mathrm { c } \\). \\( \overline { \mathrm { lk } } \cong \overline { \mathrm { np } } \\)
\\( \bigcirc \mathrm { d } \\). \\( \overline { \mathrm { lm } } \cong \overline { \mathrm { pm } } \\)

Explanation:

Step1: Recall the Angle - Side - Angle (ASA) rule

ASA states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent.

Step2: Identify the given angle and the vertical angle

We are given \(\angle L\cong\angle N\). Also, \(\angle KML\cong\angle PMN\) (vertical angles are congruent).

Step3: Determine the included side

For \(\triangle KLM\) and \(\triangle PNM\), the included side between \(\angle L\) and \(\angle KML\) is \(LM\), and the included side between \(\angle N\) and \(\angle PMN\) is \(NM\). If \(LM\cong NM\), then by ASA \(\triangle KLM\cong\triangle PNM\)

Answer:

B. \(\overline{LM}\cong\overline{NM}\)