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. \
ewline $\square$ point $p$ is the midpoint of $\overline{mk}$. \
ewline $\square$ line $mk$ is the perpendicular bisector of $\overline{ln}$. \
ewline $\square$ $\overline{ml} \cong \overline{mp}$ \
ewline $\square$ $\overline{ml} \cong \overline{mn}$ \
ewline $\square$ $\overline{pk} \cong \overline{pk}$

Explanation:

HL (Hypotenuse-Leg) congruence applies to right triangles, requiring a right angle, congruent hypotenuses, and one pair of congruent legs.

1. For $\triangle LMP$ and $\triangle NMP$ to be right triangles, $\angle LPM$ and $\angle NPM$ must be right angles. If Line $MK$ is the perpendicular bisector of $\overline{LN}$, then $MK \perp LN$, so $\angle LPM = \angle NPM = 90^\circ$, establishing the right angles.
2. $\overline{MP}$ is a shared leg (common to both triangles). For HL, we need congruent hypotenuses: $\overline{ML}$ and $\overline{MN}$ are the hypotenuses of the two right triangles, so $\overline{ML} \cong \overline{MN}$ provides the required congruent hypotenuses.
3. The other options do not satisfy HL: Point P being the midpoint only gives $\overline{LP} \cong \overline{NP}$ (not sufficient for HL), $\overline{ML} \cong \overline{MP}$ does not relate hypotenuses/legs correctly, and $\overline{PK} \cong \overline{PK}$ is irrelevant to the two target triangles.

Answer:

• Line MK is the perpendicular bisector of $\overline{LN}$.
• $\overline{ML} \cong \overline{MN}$