Question

consider $\triangle jkl$ and $\triangle mno$.

two right triangles, $\triangle jkl$ with right angle at $l$ and a marked congruent segment on $jk$, $\triangle mno$ with right angle at $o$ and a marked congruent segment on $mn$

which piece of information can be used to prove that $\triangle jkl \cong \triangle mno$?

$\bigcirc \\ \angle j \cong \angle n$

$\bigcirc \\ \angle j \cong \angle o$

$\bigcirc \\ \angle k \cong \angle m$

$\bigcirc \\ \angle k \cong \angle n$

Explanation:

Step1: Recall the Hypotenuse - Leg (HL) Congruence Theorem for right - triangles

For two right - triangles \(\triangle JKL\) and \(\triangle MNO\) (\(\angle L=\angle O = 90^{\circ}\)), if the hypotenuse (\(JK = MN\) as shown by the tick marks) and one leg are congruent, or if we can establish an Angle - Angle - Side (AAS) or Angle - Side - Angle (ASA) congruence. Since we already have the hypotenuse congruent (\(JK = MN\)) and right - angles (\(\angle L=\angle O\)), we need an additional pair of angles.

Step2: Analyze each angle - congruence option

• If \(\angle J\cong\angle N\), we have \(\angle J\cong\angle N\), \(JK = MN\), and \(\angle L=\angle O\). But for AAS, the side should be non - included.
• If \(\angle J\cong\angle O\), \(\angle O\) is a right - angle and \(\angle J\) is not a right - angle (\(\angle L\) is the right - angle in \(\triangle JKL\)), so \(\angle J

ot\cong\angle O\).

• If \(\angle K\cong\angle M\), we have \(\angle K\cong\angle M\), \(JK = MN\), and \(\angle L=\angle O\). By AAS (Angle - Angle - Side: two angles and a non - included side), \(\triangle JKL\cong\triangle MNO\).
• If \(\angle K\cong\angle N\), there is no congruence criterion (like AAS, ASA, SSS, SAS, HL) that can be applied with the given hypotenuse (\(JK = MN\)) and right - angles (\(\angle L=\angle O\)) to prove the triangles congruent.

Answer:

\(\angle K\cong\angle M\)