Answer
First, we need to prove $\triangle FED\cong\triangle GED$ by method SSS.
1. $\overline{DF}\cong\overline{DG}$ and $\overline{FE}\cong\overline{EG}$. (Given)
2. $\overline{DE}\cong\overline{DE}$ (Identity)
Then we can deduce $ \angle EDF\cong\angle EDG$. So, $\vec{DE}$ bisects $\angle FDG$
Work Step by Step
*PLANNING:
First, we need to prove $\triangle FED\cong\triangle GED$.
Then we can deduce $ \angle EDF\cong\angle EDG$. So, $\vec{DE}$ bisects $\angle FDG$
1. $\overline{DF}\cong\overline{DG}$ and $\overline{FE}\cong\overline{EG}$. (Given)
2. $\overline{DE}\cong\overline{DE}$ (Identity)
So now we have all 3 sides of $\triangle FED$ are congruent with 3 corresponding sides of $\triangle GED$. Therefore,
3. $\triangle FED\cong\triangle GED$ (SSS)
4. $\angle EDF\cong\angle EDG$ (CPCTC)
5. $\vec{DE}$ bisects $\angle FDG$ (if an angle is divided into 2 congruent angles by a line, then that line bisects that angle)
