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]
### 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]
