Question

step\tstatement\treason
1\tabcd is a rhombus
\t\overline{df} bisects \overline{bc}\tgiven
2\t\overline{be} \cong \overline{ce}\ta segment bisector divides a segment into two congruent segments
3\t\angle dec \cong \angle feb\tvertical angles are congruent
4\t\overline{dc} \parallel \overline{ab}\topposite sides of a parallelogram / rhombus / rectangle / square are parallel
5\t\angle c \cong \angle ebf\tparallel lines cut by a transversal form congruent alternate interior angles
6\t\triangle dec \cong \triangle feb\t(select a reason)
(image of a rhombus abcd with points d, c, a, b, f and e as described in the statements)

Explanation:

Step1: Identify congruent parts

We have $\overline{BE} \cong \overline{CE}$ (from step 2), $\angle DEC \cong \angle FEB$ (from step 3), and $\angle C \cong \angle EBF$ (from step 5).

Step2: Determine congruence criterion

The congruence criterion that uses two angles and the included side is ASA (Angle - Side - Angle). Here, $\angle DEC \cong \angle FEB$ (angle), $\overline{CE} \cong \overline{BE}$ (side, since $\overline{BE} \cong \overline{CE}$), and $\angle C \cong \angle EBF$ (angle). So the reason for $\triangle DEC \cong \triangle FEB$ is ASA (Angle - Side - Angle) congruence criterion.

Answer:

The reason for $\triangle DEC \cong \triangle FEB$ is "ASA (Angle - Side - Angle) congruence criterion" (or more precisely, "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" which is the ASA postulate).