Question

given: abcd is a parallelogram and e is the midpoint of \\(\overline{bc}\\).
prove: \\(\triangle fbe \cong \triangle dce\\).

step	statement	reason
2	\\(\overline{be} \cong \overline{ec}\\)	a midpoint divides a segment into two congruent segments
3	\\(\angle dec \cong \angle feb\\)	vertical angles are congruent
4	\\(dc \parallel ab\\)	opposite sides of a parallelogram / rhombus / rectangle / square are parallel
5	\\(\angle c \cong \angle ebf\\)	parallel lines cut by a transversal form congruent alternate interior angles
6	\\(\triangle fbe \cong \triangle dce\\)	select a reason...

(note: df, af and bc are segments)

answer

Explanation:

Step1: Identify congruent parts

We have \( \overline{BE} \cong \overline{EC} \) (from midpoint), \( \angle DEC \cong \angle FEB \) (vertical angles), and \( \angle C \cong \angle EBF \) (alternate interior angles from parallel lines \( DC \parallel AB \)).

Step2: Apply ASA congruence

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 two triangles are congruent. In \( \triangle FBE \) and \( \triangle DCE \):

• \( \angle EBF \cong \angle C \) (alternate interior angles)
• \( \overline{BE} \cong \overline{EC} \) (midpoint property)
• \( \angle FEB \cong \angle DEC \) (vertical angles)

So, by the ASA (Angle - Side - Angle) congruence postulate, \( \triangle FBE \cong \triangle DCE \).

Answer:

The reason for \( \triangle FBE \cong \triangle DCE \) is "ASA (Angle - Side - Angle) Congruence Postulate" (or "Angle - Side - Angle congruence criterion")