Question

step
statement\treason
1\tabcd is a parallelogram
\t\overline{bg} \cong \overline{dh}
\t\overline{ae} \cong \overline{fc}\tgiven
2\t\overline{ef} \cong \overline{ef}\treflexive property
3\t\overline{af} \cong \overline{ce}\tcongruent segments added to congruent segments form congruent segments
4\t\overline{ad} \cong \overline{bc}\topposite sides of a parallelogram / rhombus / rectangle / square are congruent
5\t\overline{ah} \cong \overline{cg}\tcongruent segments subtracted from congruent segments form congruent segments
6\t\overline{ad} \parallel \overline{bc}\topposite sides of a parallelogram / rhombus / rectangle / square are parallel
7\t\angle gce \cong \angle haf\tselect a reason...

(note: there is a diagram at the bottom with parallelogram abcd, points e, f, g, h on various sides, and segments marked with congruency and parallelism indicators. also, note: \overline{ac}, \overline{bc} and \overline{ad} are segments.)

Explanation:

Step1: Recall properties of parallel lines

Since \(AD \parallel BC\) (from step 6, opposite sides of a parallelogram are parallel), and \(AC\) is a transversal.

Step2: Apply Alternate Interior Angles Theorem

When two parallel lines (\(AD\) and \(BC\)) are cut by a transversal (\(AC\)), the alternate interior angles are congruent. So, \(\angle GCE\) and \(\angle HAF\) are alternate interior angles formed by transversal \(AC\) with parallel lines \(AD\) and \(BC\), hence \(\angle GCE \cong \angle HAF\) by the Alternate Interior Angles Theorem.

Answer:

Alternate Interior Angles Theorem (or: If two parallel lines are cut by a transversal, then alternate interior angles are congruent)