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. \\( \square \angle zwy \cong \angle xyw \\) by the alternate interior \\( \angle s \\) theorem. \\( \square \overline{wy} \cong \overline{wy} \\) by the reflexive property. \\( \square \angle zwy \cong \angle xwy \\) by the corresponding \\( \angle s \\) theorem. \\( \square \overline{wx} \cong \overline{zy} \\) by definition of a parallelogram. \\( \square \overline{wz} \cong \overline{xy} \\) by the given.

Explanation:

• For $\angle ZWY\cong\angle XYW$: If $WZ\parallel XY$ (which is a property we might assume when proving a parallelogram, and by the alternate - interior angles theorem (when a transversal $WY$ intersects two parallel lines $WZ$ and $XY$), $\angle ZWY$ and $\angle XYW$ are congruent.
• For $\overline{WY}\cong\overline{WY}$: By the reflexive property of congruence (a segment is congruent to itself).
• For $\overline{WX}\cong\overline{ZY}$: In a parallelogram, opposite sides are congruent.
• For $\overline{WZ}\cong\overline{XY}$: This is given (as indicated by the tick - marks on the sides in the diagram).

• $\angle ZWY\cong\angle XWY$ is incorrect. $\angle ZWY$ and $\angle XWY$ are not corresponding angles in the context of proving the two triangles congruent for the parallelogram proof.

Answer:

$\angle ZWY\cong\angle XYW$ by the alternate interior $\angle$s theorem, $\overline{WY}\cong\overline{WY}$ by the reflexive property, $\overline{WX}\cong\overline{ZY}$ by definition of a parallelogram, $\overline{WZ}\cong\overline{XY}$ by the given.