Question

there is a right triangle x y i with right angle at i, and another right triangle m n z with right angle at n. point z is on the line segment i y (or its extension) and on the line segment x m. on the right side, there are two radio buttons: one labeled not similar or not necessarily similar and the other labeled similar:. below the similar: option, there is a statement △xyz ~ △blank by the followed by a dropdown menu labeled select.

Explanation:

Step1: Identify Angles

$\angle XYZ$ and $\angle MZN$ are vertical angles, so $\angle XYZ = \angle MZN$. Also, $\angle XJY = \angle MNY = 90^\circ$ (right angles).

Step2: Apply AA Similarity

By the Angle - Angle (AA) similarity criterion, if two angles of one triangle are equal to two angles of another triangle, the triangles are similar. So $\triangle XJY \sim \triangle MNY$? Wait, no, looking at the triangles: $\triangle XYZ$ (wait, maybe $\triangle XJY$ and $\triangle MNY$? Wait, the right angles are $\angle J$ and $\angle N$. So $\angle J = \angle N = 90^\circ$, and $\angle XZJ=\angle MZN$ (vertical angles). So $\triangle XJY \sim \triangle MNY$ by AA? Wait, the problem has $\triangle XYZ - \triangle$ [blank]. Wait, maybe $\triangle XJY$ and $\triangle MNY$? Wait, the labels: $J$, $X$, $Y$ and $N$, $M$, $Z$. So $\angle J = \angle N = 90^\circ$, $\angle XZY=\angle MZN$ (vertical angles), so $\triangle XJY \sim \triangle MNY$ by AA similarity. Wait, the first triangle is $\triangle XYZ$? Wait, no, the right angle is at $J$ for the left triangle, so the left triangle is $\triangle XJY$ (right - angled at $J$), and the right triangle is $\triangle MNY$ (right - angled at $N$). So the two triangles $\triangle XJY$ and $\triangle MNY$ are similar by AA (Angle - Angle) similarity, since two angles (right angle and vertical angle) are equal.

Answer:

Similar: $\triangle XYZ \sim \triangle MZY$ (wait, maybe $\triangle XJY \sim \triangle MNY$) by the AA (Angle - Angle) Similarity Criterion. Wait, the correct triangles: $\triangle XJY$ (right - angled at $J$) and $\triangle MNY$ (right - angled at $N$), with $\angle XZY=\angle MZN$ (vertical angles) and $\angle J=\angle N = 90^\circ$. So the triangles are similar by AA, so the answer is "Similar: $\triangle XJY \sim \triangle MNY$ by the AA (Angle - Angle) Similarity" (assuming the triangles are $\triangle XJY$ and $\triangle MNY$). Wait, the first triangle is $\triangle XYZ$? Maybe a typo, but the key is AA similarity. So the answer is that the triangles are similar by AA, so we select "Similar" and the similarity criterion is AA.