proof:
1. quadrilateral defg is a parallelogram. - given
2. $ \overline { d e } \parallel \overline { g f } $ $ \overline { d g } \parallel \overline { e f } $ - definition of a parallelogram
3. draw $ \overline { d f } $ and $ \overline { g e } $. these line segments are transversals cutting two pairs of parallel lines: $ \overleftrightarrow { d e } $ and $ \overleftrightarrow { g f } $ and $ \overleftrightarrow { d g } $ and $ \overleftrightarrow { e f } $. - drawing line segments
4. place point h where $ \overline { d f } $ and $ \overline { g e } $ intersect. - defining a point
5. $ \angle h g d \cong \angle h e f $ $ \angle h d g \cong \angle h f e $ -
6. $ \overline { d g } \cong \overline { e f } $ - opposite sides of a parallelogram are congruent.
7. - asa criterion for congruence
8. $ \overline { g h } \cong \overline { e h } $ $ \overline { d h } \cong \overline { f h } $ - corresponding sides of congruent triangles are congruent.
10
what is the reason for step 5 in this proof?
a. vertical angles theorem
b. opposite interior angles of a parallelogram are congruent.
c. all angles of a rectangle have the same measure.
d. alternate interior angles theorem
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# Brief Explanations:
- **Vertical Angles Theorem**: States that vertical angles (opposite angles formed by two intersecting lines) are congruent. But in this case, the angles in step 5 are not vertical angles.
- **Opposite interior angles of a parallelogram are congruent**: This is not a standard theorem. The standard properties are about opposite sides and opposite angles (not “opposite interior angles” in this context) of a parallelogram.
- **All angles of a rectangle have the same measure**: A rectangle is a special - case of a parallelogram, but we are dealing with a general parallelogram $DEFG$ here, not specifically a rectangle.
- **Alternate Interior Angles Theorem**: When two parallel lines ($\overline{DG}\parallel\overline{EF}$ as per the definition of a parallelogram in step 2) are cut by a transversal ($\overline{GE}$ for $\angle HGD$ and $\angle HEF$, and $\overline{DF}$ for $\angle HDG$ and $\angle HFE$), the alternate interior angles are congruent.

# Answer:
D. Alternate Interior Angles Theorem

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