Question

examine this figure. which two pieces of information, if true, would help to prove that \\( \triangle lmp \cong \triangle nmp \\) by hl? select two options. \\( \square \\) point \\( p \\) is the midpoint of \\( \overline{mk} \\). \\( \square \\) line \\( mk \\) is the perpendicular bisector of \\( \overline{ln} \\). \\( \square \overline{ml} \cong \overline{mp} \\) \\( \square \overline{ml} \cong \overline{mn} \\) \\( \square \overline{pk} \cong \overline{pk} \\)

Explanation:

Step1: Recall HL (Hypotenuse - Leg) theorem

HL theorem states that if the hypotenuse and a leg of a right - triangle are congruent to the hypotenuse and a leg of another right - triangle, then the two right - triangles are congruent.

Step2: Analyze each option

• Option 1: Point \(P\) is the midpoint of \(\overline{MK}\)

This information is not relevant for proving \(\triangle LMP\cong\triangle NMP\) by HL.

• Option 2: Line \(MK\) is the perpendicular bisector of \(\overline{LN}\)

If \(MK\) is the perpendicular bisector of \(\overline{LN}\), then \(\angle LPM=\angle NPM = 90^{\circ}\) (right angles). Also, \(MP = MP\) (common side). If we can get the hypotenuse congruent, we can use HL.

• Option 3: \(\overline{ML}\cong\overline{MP}\)

This gives a leg - leg (not hypotenuse - leg) situation for right - triangles \(\triangle LMP\) and \(\triangle NMP\) (if they are right - triangles).

• Option 4: \(\overline{ML}\cong\overline{MN}\)

If \(\overline{ML}\cong\overline{MN}\) (hypotenuse) and since \(MP = MP\) (common leg) and \(\angle LPM=\angle NPM = 90^{\circ}\) (from the perpendicular bisector property in option 2), then by HL \(\triangle LMP\cong\triangle NMP\).

• Option 5: \(\overline{PK}\cong\overline{PK}\)

This is a reflexive property but not relevant for proving \(\triangle LMP\cong\triangle NMP\) by HL.

Answer:

Line \(MK\) is the perpendicular bisector of \(\overline{LN}\), \(\overline{ML}\cong\overline{MN}\)