Question

to prove quadrilateral wxyz is a parallelogram, travis begins by proving $\triangle wzy \cong \triangle yxw$ by using the sas congruency theorem.
which reasons can travis use to prove the two triangles are congruent? check all that apply.
$\angle zwy \cong \angle xyw$ by the alternate interior $\angle$s theorem.
$\overline{wy} \cong \overline{wy}$ by the reflexive property.
$\angle zwy \cong \angle xwy$ by the corresponding $\angle$s theorem.
$\overline{wx} \cong \overline{zy}$ by definition of a parallelogram.
$\overline{wz} \cong \overline{xy}$ by the given.

Explanation:

1. For \( \angle ZWY \cong \angle XYW \) by alternate interior \( \angle s \) theorem: If \( WZ \parallel XY \) (implied by markings or parallelogram attempt), then transversal \( WY \) creates alternate interior angles \( \angle ZWY \) and \( \angle XYW \), so this is valid.
2. For \( \overline{WY} \cong \overline{WY} \) by reflexive property: A segment is congruent to itself, so this is valid for SAS (side - angle - side, the common side).
3. \( \angle ZWY \cong \angle XWY \) by corresponding \( \angle s \) theorem: There's no corresponding angles setup here (corresponding angles are for parallel lines cut by transversal with same - position angles), so this is invalid.
4. \( \overline{WX} \cong \overline{ZY} \) by definition of parallelogram: We are trying to prove it's a parallelogram, so we can't use the definition as a reason before proving, so invalid.
5. \( \overline{WZ} \cong \overline{XY} \) by the given: The markings on \( WZ \) and \( XY \) suggest they are given as congruent, so this is valid (for SAS, a side).

Answer:

• \( \angle ZWY \cong \angle XYW \) by the alternate interior \( \angle s \) theorem.
• \( \overline{WY} \cong \overline{WY} \) by the reflexive property.
• \( \overline{WZ} \cong \overline{XY} \) by the given.