Question

1. examine the given figure. ( overline{np} ) bisects ( overline{lm} ) and ( angle m cong angle l ). if ( overline{np} ) is a perpendicular bisector to ( overline{lm} ), then ______________ ( \triangle mpn cong \triangle lpn ) by asa ( \triangle mpn cong \triangle lpn ) by sss ( \triangle mpn cong \triangle lpn ) by aas ( \triangle mpn cong \triangle lpn ) by ssa

Explanation:

Step1: Analyze the given information

Since \(NP\) is a perpendicular bisector of \(LM\), we have \(MP = LP\) (definition of perpendicular bisector). Also, \(\angle MPN=\angle LPN = 90^{\circ}\) (perpendicular bisector property), and \(NP\) is common to both \(\triangle MPN\) and \(\triangle LPN\) (\(NP = NP\) by reflexive property).

Step2: Recall triangle - congruence criteria

• ASA (Angle - Side - Angle): Two angles and the included side of one triangle are equal to two angles and the included side of another triangle.
• SSS (Side - Side - Side): All three sides of one triangle are equal to all three sides of another triangle.
• AAS (Angle - Angle - Side): Two angles and a non - included side of one triangle are equal to two angles and the corresponding non - included side of another triangle.
• SSA (Side - Side - Angle): Not a valid congruence criterion in general (except for right - angled triangles in a special case, but not a standard congruence rule like ASA, SSS, AAS).

We know \(\angle M\cong\angle L\), \(\angle MPN\cong\angle LPN\), and \(NP = NP\). The side \(NP\) is not between \(\angle M\) and \(\angle L\). But we have two angles (\(\angle M\cong\angle L\) and \(\angle MPN\cong\angle LPN\)) and a non - included side (\(NP\))

Answer:

\(\triangle MPN\cong\triangle LPN\) by AAS.