Question

given: \\( \triangle dfe \\) is isosceles with base \\( \overline{fe} \\); \\( \overline{fb} \cong \overline{ec} \\). prove: \\( \triangle dfb \cong \triangle dec \\) complete the missing parts of the paragraph proof. we know that triangle dfe is isosceles with base \\( \overline{fe} \\) and that segment fb is congruent to segment ec because. segment df is congruent to segment by the definition of isosceles triangle. since these segments are congruent, the base angles, angles, are congruent by the isosceles triangle theorem. therefore, triangles are congruent by sas.

Explanation:

1. The first blank is filled with "it is given" as the problem statement provides that \( \overline{FB}\cong\overline{EC} \).
2. In an isosceles triangle \( \triangle DFE \) with base \( \overline{FE} \), the non - base sides \( \overline{DF} \) and \( \overline{DE} \) are congruent by the definition of an isosceles triangle.
3. The base angles of an isosceles triangle \( \triangle DFE \) (angles \( \angle F \) and \( \angle E \)) are congruent by the isosceles triangle theorem.
4. For the SAS (Side - Angle - Side) congruence criterion, in \( \triangle DFB \) and \( \triangle DEC \), we have \( \overline{FB}\cong\overline{EC} \) (given), \( \overline{DF}\cong\overline{DE} \) (from the isosceles triangle \( \triangle DFE \)), and \( \angle F\cong\angle E \) (base angles of \( \triangle DFE \)). So, \( \triangle DFB \) and \( \triangle DEC \) are congruent by SAS.

Answer:

1. it is given
2. \( DE \)
3. \( \angle F \) and \( \angle E \)
4. \( \triangle DFB \) and \( \triangle DEC \)