Question

given: abcd is a parallelogram, ∠gec ≅ ∠hfa and \\(\overline{ae} \cong \overline{fc}\\).
prove: \\(\triangle gec \cong \triangle hfa\\).
step\tstatement\treason
1\tabcd is a parallelogram
\\(\angle gec \cong \angle hfa\\)
\\(\overline{ae} \cong \overline{fc}\\)\tgiven
2\t\\(\overline{ef} \cong \overline{ef}\\)\tselect a reason

Explanation:

Step1: Identify the property for \( \overline{EF} \cong \overline{EF} \)

The segment \( \overline{EF} \) is congruent to itself. This is the Reflexive Property of Congruence, which states that any geometric figure is congruent to itself. So the reason for \( \overline{EF} \cong \overline{EF} \) is the Reflexive Property of Congruence.

Step2: Analyze \( AE \cong FC \) and \( EF \cong EF \) to get \( EC \cong FA \)

We know that \( AE \cong FC \) (given) and \( EF \cong EF \) (from Step 1). By the Segment Addition Postulate, \( EC = EF + FC \) and \( FA = EF + AE \). Since \( AE \cong FC \) (so \( AE = FC \)) and \( EF = EF \), we can substitute to get \( EC = FA \), which means \( \overline{EC} \cong \overline{FA} \) (by the definition of congruent segments, if two segments have the same length, they are congruent).

Step3: Prove \( \triangle GEC \cong \triangle HFA \)

We have \( \angle GEC \cong \angle HFA \) (given), \( \overline{EC} \cong \overline{FA} \) (from Step 2), and we can check the angles or sides. Wait, actually, with \( \angle GEC \cong \angle HFA \), \( \overline{EC} \cong \overline{FA} \), and we can also note that maybe \( \angle ECG \cong \angle FAH \) because \( ABCD \) is a parallelogram (so \( AB \parallel CD \), leading to alternate interior angles being congruent). But more directly, with \( \angle GEC \cong \angle HFA \), \( \overline{EC} \cong \overline{FA} \), and if we consider the ASA (Angle - Side - Angle) or SAS (Side - Angle - Side) criterion. Wait, we have \( \angle GEC \cong \angle HFA \) (angle), \( \overline{EC} \cong \overline{FA} \) (side), and we can get another angle. Since \( ABCD \) is a parallelogram, \( AB \parallel CD \), so \( \angle ECD \cong \angle FAB \) (alternate interior angles). But maybe more simply, with \( \angle GEC \cong \angle HFA \), \( \overline{EC} \cong \overline{FA} \), and if we assume the included angles or another side. Wait, actually, we have:

• \( \angle GEC \cong \angle HFA \) (given)
• \( \overline{EC} \cong \overline{FA} \) (proven)
• We can also note that \( \angle ECG \cong \angle FAH \) (because \( ABCD \) is a parallelogram, \( AB \parallel CD \), so alternate interior angles are congruent). But actually, with \( \angle GEC \cong \angle HFA \), \( \overline{EC} \cong \overline{FA} \), and \( \angle ECG \cong \angle FAH \), we can use ASA. Or with \( \angle GEC \cong \angle HFA \), \( \overline{EC} \cong \overline{FA} \), and if we consider the other side, but let's focus on the steps.

First, the key steps for the proof:

1. \( ABCD \) is a parallelogram, \( \angle GEC \cong \angle HFA \), \( \overline{AE} \cong \overline{FC} \) (Given)
2. \( \overline{EF} \cong \overline{EF} \) (Reflexive Property of Congruence)
3. \( \overline{EC} \cong \overline{FA} \) (Segment Addition Postulate and definition of congruent segments, since \( EC=EF + FC\) and \( FA = EF+AE\), and \( AE = FC\))
4. \( \triangle GEC \cong \triangle HFA \) (ASA or SAS: we have \( \angle GEC \cong \angle HFA \), \( \overline{EC} \cong \overline{FA} \), and if we take \( \angle ECG \cong \angle FAH \) (from parallelogram \( ABCD \), \( AB \parallel CD \), alternate interior angles), so by ASA, \( \triangle GEC \cong \triangle HFA \))

But the main part for the given step (step 2 in the table) is that \( \overline{EF} \cong \overline{EF} \) is by the Reflexive Property of Congruence.

Answer:

The reason for \( \overline{EF} \cong \overline{EF} \) is the Reflexive Property of Congruence. And the proof of \( \triangle GEC \cong \triangle HFA \) can be done using the ASA (or SAS) congruence criterion with the established congruent angles and sides.