Question

prove: $\triangle gec \cong \triangle hfa$.
step\tstatement\treason
1\t$abcd$ is a parallelogram
$\angle gec \cong \angle hfa$
$\overline{ae} \cong \overline{fc}$\tgiven
2\t$\overline{ef} \cong \overline{ef}$\treflexive property
3\t$\overline{af} \cong \overline{ce}$\tsubstitution
4\t$\overline{bc} \parallel \overline{ad}$\topposite sides of a parallelogram / rhombus / rectangle / square are parallel
5\t$\angle gce \cong \angle fah$\tparallel lines cut by a transversal form congruent alternate interior angles
6\t$\triangle gec \cong \triangle hfa$\tblank
image of a parallelogram with points b, g, c, f, e, a and lines

Explanation:

Step1: Identify congruent parts

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

Step2: Apply AAS congruence

The Angle - Angle - Side (AAS) congruence criterion states that if two angles and a non - included side of one triangle are congruent to the corresponding two angles and non - included side of another triangle, then the two triangles are congruent. For $\triangle GEC$ and $\triangle HFA$, $\angle GEC \cong \angle HFA$, $\angle GCE \cong \angle FAH$, and $\overline{EC} \cong \overline{FA}$ (which is the non - included side between the two angles). So by AAS congruence criterion, $\triangle GEC \cong \triangle HFA$.

Answer:

The reason for $\triangle GEC \cong \triangle HFA$ is the Angle - Angle - Side (AAS) Congruence Theorem.