Question

given \\( \triangle dfe \\) is isosceles with base fe. \\( fb = 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 \\( \sqrt { } \\) segment df is congruent to segment \\( \sqrt { } \\) by the definition of isosceles triangle. since these segments are congruent, the base angles, angles \\( \sqrt { } \\) are congruent by the isosceles triangle theorem. therefore, triangles \\( \sqrt { } \\) are congruent by sas

Explanation:

Step1: Use given information

We are given \(FB = EC\).

Step2: Use isosceles triangle property

In isosceles \(\triangle DFE\) with base \(FE\), \(DF = DE\) (by definition of isosceles triangle).

Step3: Use base - angles of isosceles triangle

The base - angles of isosceles \(\triangle DFE\) are \(\angle F\) and \(\angle E\), so \(\angle F=\angle E\) (by the base - angles of isosceles triangle theorem).

Step4: Apply SAS congruence

For \(\triangle DFB\) and \(\triangle DEC\), we have \(DF = DE\), \(\angle F=\angle E\), and \(FB = EC\). So, \(\triangle DFB\cong\triangle DEC\) (by SAS congruence).

Answer:

First blank: \(FB = EC\) (given). Second blank: \(DE\). Third blank: \(\angle F\) and \(\angle E\). Fourth blank: \(\triangle DFB\) and \(\triangle DEC\)