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
∠gec ≅ ∠hfa
\\(\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{bc} \parallel \overline{ad}\\)\topposite sides of a parallelogram / rhombus / rectangle / square are parallel
5\t∠gce ≅ ∠fah\tparallel lines cut by a transversal form congruent alternate interior angles
6\t\\(\triangle gec \cong \triangle hfa\\)\tselect a reason...
(note: \\(\overline{ac}\\), \\(\overline{bc}\\) and \\(\overline{ad}\\) are segments.)

Explanation:

Step1: Identify known congruences

We know \( \angle GEC \cong \angle HFA \) (given), \( \overline{AF} \cong \overline{CE} \) (from step 3), and \( \angle GCE \cong \angle FAH \) (from step 5).

Step2: Determine triangle congruence criterion

The Angle - Side - Angle (ASA) congruence criterion states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent. Here, \( \angle GEC \cong \angle HFA \), \( \overline{CE} \cong \overline{AF} \), and \( \angle GCE \cong \angle FAH \), so by ASA, \( \triangle GEC \cong \triangle HFA \).

Answer:

The reason is "Angle - Side - Angle (ASA) Congruence Criterion" (or simply "ASA" as the reason for triangle congruence).