Question

statement reason
abcd is a parallelogram
e is the midpoint of \\(\overline{ab}\\) given
f is the midpoint of \\(\overline{dc}\\)
\\(\overline{ad} \cong \overline{cb}\\) opposite sides of a parallelogram / rhombus / rectangle / square are congruent
\\(\overline{dc} \cong \overline{ab}\\) opposite sides of a parallelogram / rhombus / rectangle / square are congruent
\\(fc = \frac{1}{2}dc\\) when a midpoint divides a segment into two parts, each part is half the length of the whole segment
\\(ae = \frac{1}{2}ab\\) when a midpoint divides a segment into two parts, each part is half the length of the whole segment
\\(\overline{fc} \cong \overline{ae}\\) segments that are half the lengths of congruent segments are congruent
\\(\angle a \cong \angle c\\) opposite angles of a parallelogram / rhombus / rectangle / square are congruent
\\(\triangle ade \cong \triangle cbf\\) select a reason...
diagram of parallelogram abcd with e midpoint of ab, f midpoint of dc, and triangles ade and cbf

Explanation:

Step1: Identify known congruences

We know \( \overline{AD} \cong \overline{CB} \), \( \overline{FC} \cong \overline{AE} \), and \( \angle A \cong \angle C \) from previous statements and reasons.

Step2: Determine triangle congruence criterion

For \( \triangle ADE \) and \( \triangle CBF \), we have two sides and the included angle congruent (SAS: Side - Angle - Side). The sides \( \overline{AD} \cong \overline{CB} \), \( \overline{AE} \cong \overline{FC} \), and the included angle \( \angle A \cong \angle C \). So the reason is "If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent (SAS)".

Answer:

If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent (SAS)