Question

which statements must be true about the reflection of
\\( \triangle x y z \\) across \\( \overleftrightarrow{m n} \\)? select three options
\\( \square m \angle x^{prime} z^{prime} y^{prime}=90^{circ} \\)
\\( \square m \angle m c y=90^{circ} \\)
\\( \square \overline{x x^{prime}} \cong \overline{y y^{prime}} \\)
\\( \square \overline{b z^{prime}} \cong \overline{b z} \\)
\\( \square \overline{x y} \\| \overline{x^{prime} y^{prime}} \\)

Explanation:

Step1: Properties of reflection

In a reflection, the line of reflection ($\overleftrightarrow{MN}$) is the perpendicular bisector of the segments joining corresponding points. So, for any point $P$ and its image $P'$ after reflection over $\overleftrightarrow{MN}$, $PP'$ is perpendicular to $\overleftrightarrow{MN}$ and the mid - point of $PP'$ lies on $\overleftrightarrow{MN}$.

Step2: Analyze $\overline{XX'}\cong\overline{YY'}$

Since $X$ and $X'$ are reflection images and $Y$ and $Y'$ are reflection images. The distance from $X$ to the line of reflection $\overleftrightarrow{MN}$ is equal to the distance from $X'$ to $\overleftrightarrow{MN}$, and the distance from $Y$ to the line of reflection $\overleftrightarrow{MN}$ is equal to the distance from $Y'$ to $\overleftrightarrow{MN}$. Also, the transformation is a rigid motion (reflection is a rigid motion), so $XX'$ and $YY'$ are both twice the distance from their respective points to the line of reflection. So, $\overline{XX'}\cong\overline{YY'}$.

Step3: Analyze $\overline{BZ'}\cong\overline{BZ}$

Since $B$ lies on the line of reflection $\overleftrightarrow{MN}$, and $Z$ and $Z'$ are reflection images. The line of reflection is the perpendicular bisector of $ZZ'$. If a point $B$ is on the line of reflection, then $BZ = BZ'$ (by the property of the perpendicular bisector: any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment). So, $\overline{BZ'}\cong\overline{BZ}$.

Step4: Analyze $\overline{XY}\parallel\overline{X'Y'}$

Reflection is a rigid motion. Rigid motions (reflections, translations, rotations) preserve parallelism. The pre - image segment $XY$ and its image segment $X'Y'$ are such that the transformation (reflection) does not change the direction (in terms of slope for a coordinate - plane interpretation, but more generally, in a plane geometric sense). So, $\overline{XY}\parallel\overline{X'Y'}$.

Step5: Analyze $m\angle X'Z'Y'=90^{\circ}$

There is no information given that $\angle XZY = 90^{\circ}$ in the original triangle $\triangle XYZ$. Reflection preserves angles. If $\angle XZY
eq90^{\circ}$, then $m\angle X'Z'Y'
eq90^{\circ}$.

Step6: Analyze $m\angle MCY = 90^{\circ}$

There is no indication from the problem statement or the general properties of reflection (unless $\overleftrightarrow{MN}\perp CY$ which is not given) that $m\angle MCY = 90^{\circ}$. Reflection does not imply this angle measure unless there is prior information about perpendicular lines in the pre - image related to the line of reflection.

Answer:

$\overline{XX'}\cong\overline{YY'}$, $\overline{BZ'}\cong\overline{BZ}$, $\overline{XY}\parallel\overline{X'Y'}$