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:

• The first blank is "Given" as \(FB\cong EC\) is provided in the problem statement.
• In an isosceles triangle \(DFE\) with base \(FE\), by the definition of an isosceles triangle, the non - base sides \(DF\) and \(DE\) are congruent.
• The base angles of an isosceles triangle \(DFE\) (with base \(FE\)) are \(\angle DFC\) and \(\angle DEB\) (angles opposite the congruent sides \(DF\) and \(DE\)).
• We want to prove \(\triangle DFB\cong\triangle DEC\). We have \(DF = DE\) (from the isosceles triangle definition), \(\angle DFB=\angle DEC\) (base angles of isosceles \(\triangle DFE\)), and \(FB = EC\) (given). So, by the Side - Angle - Side (SAS) congruence criterion, \(\triangle DFB\cong\triangle DEC\).

Answer:

1. Given
2. \(DE\)
3. \(\angle DFC\) and \(\angle DEB\)
4. \(DFB\) and \(DEC\)