[tex]$\overline{YS}$[/tex] is the perpendicular bisector of [tex]$\Delta XYZ$[/tex], and [tex]$\overline{YS}$[/tex] is a shared side of [tex]$\triangle XYS$[/tex] and [tex]$\triangle ZYS$[/tex]. Which of the following must be congruent in order to verify that [tex]$\triangle XYS \cong \triangle ZYS$[/tex]?

A. [tex]$\overline{ZY} \cong \overline{ZS}$[/tex]
B. [tex]$\overline{XY} \cong \overline{YS}$[/tex]
C. [tex]$\overline{XS} \cong \overline{ZS}$[/tex]
D. [tex]$\overline{XS} \cong \overline{XY}$[/tex]



Answer :

To verify that [tex]\(\triangle XYS \cong \triangle ZYS\)[/tex], we need to use congruence criteria such as SAS (Side-Angle-Side), ASA (Angle-Side-Angle), or SSS (Side-Side-Side). Since [tex]\(\overline{YS}\)[/tex] is already given to be a shared side, we need to detail further steps to check for congruence.

### Given Information:
- [tex]\(\overline{YS}\)[/tex] is the perpendicular bisector of [tex]\(\overline{XZ}\)[/tex].
- [tex]\(\overline{YS}\)[/tex] is a common side of both triangles [tex]\(\triangle XYS\)[/tex] and [tex]\(\triangle ZYS\)[/tex].

### Steps to Solve:

1. Perpendicular Bisector Property:
Since [tex]\(\overline{YS}\)[/tex] is the perpendicular bisector of [tex]\(\overline{XZ}\)[/tex], it implies two things:
- [tex]\(\overline{XS} \cong \overline{ZS}\)[/tex]
- [tex]\(\angle YSX \cong \angle YSZ = 90^\circ\)[/tex]

2. Common Side:
Both triangles share side [tex]\(\overline{YS}\)[/tex].

3. Congruence Criteria:
To show [tex]\(\triangle XYS \cong \triangle ZYS\)[/tex], we consider the Side-Angle-Side (SAS) criterion:
- Side 1: [tex]\(\overline{YS}\)[/tex] (common to both triangles)
- Angle: [tex]\(\angle YSX \cong \angle YSZ\)[/tex] (both right angles from the perpendicular bisector)
- Side 2: [tex]\(\overline{XS} \cong \overline{ZS}\)[/tex] (since [tex]\(\overline{YS}\)[/tex] is the perpendicular bisector of [tex]\(\overline{XZ}\)[/tex])

From the given options:

- Option A: [tex]\(\overline{ZY} \cong \overline{ZS}\)[/tex] is not implied by the given conditions.
- Option B: [tex]\(\overline{XY} \cong \overline{YS}\)[/tex] is not necessarily true and not required by the given conditions.
- Option C: [tex]\(\overline{XS} \cong \overline{ZS}\)[/tex] is true as [tex]\(\overline{YS}\)[/tex] is the perpendicular bisector of [tex]\(\overline{XZ}\)[/tex].
- Option D: [tex]\(\overline{XS} \simeq \overline{XY}\)[/tex] is irrelevant and incorrect.

### Answer:
The sides that must be congruent in order to verify that [tex]\(\triangle XYS \cong \triangle ZYS\)[/tex] are:
[tex]\[ \boxed{\overline{X S} \cong \overline{Z S}}. \][/tex]