Question

proving triangles congruent
triangle congruence: sas
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 fe and that segment fb is congruent to segment ec because
segment df is congruent to segment \\( \checkmark \\) by the definition of isosceles triangle. since these
segments are congruent, the base angles \\( \checkmark \\) are congruent by the isosceles triangle theorem. therefore, triangles
\\( \checkmark \\) are congruent by sas

Explanation:

Step1: Identify congruent sides and angles

Since \(\triangle DFE\) is isosceles with base \(FE\), by the definition of isosceles triangle, \(DF = DE\). Given \(FB\cong EC\). Also, \(\angle DFB\) and \(\angle DEC\) are congruent because the base angles of an isosceles triangle (\(\triangle DFE\)) are congruent (by the isosceles triangle theorem).

Step2: Apply SAS congruence criterion

In \(\triangle DFB\) and \(\triangle DEC\), we have \(DF = DE\) (from isosceles triangle definition), \(FB\cong EC\) (given), and \(\angle DFB\cong\angle DEC\) (isosceles triangle base - angles). So, by the Side - Angle - Side (SAS) congruence criterion, \(\triangle DFB\cong\triangle DEC\).

Answer:

Segments \(DF\) and \(DE\) are congruent by the definition of isosceles triangle. Angles \(\angle DFB\) and \(\angle DEC\) are congruent by the isosceles triangle theorem. Triangles \(\triangle DFB\) and \(\triangle DEC\) are congruent by SAS.