Chapter 2.6, Problem 4PSA

To determine

To Prove : $\angle 1\cong \angle 2$

Explanation of Solution

Given information: $\angle AFE\cong \angle DEF$

  

  $\because \overline{FC}$ Bisects $\angle AFE$

  $\therefore \angle AFC=\angle CFE$

Similarly $\angle DEB=\angle BEF$

  $\begin{array}{l}\because \angle AFE\cong \angle DEF\\ \cancel{2}\angle 1=\cancel{2}\angle 2\\ \angle 1=\angle 2\end{array}$