Answer
First, we need to prove $\triangle FED\cong\triangle GED$ by SSS.
Then we can deduce $\angle DEF\cong\angle DEG$.
Then prove $\angle DEF$ and $\angle DEG$ must be right $\angle$s.
That means $\overline{DE}\bot\overline{FG}$
Work Step by Step
*PLANNING:
First, we need to prove $\triangle FED\cong\triangle GED$.
Then we can deduce $\angle DEF\cong\angle DEG$.
So $\angle DEF$ and $\angle DEG$ must be right $\angle$s.
That means $\overline{DE}\bot\overline{FG}$
1. $E$ is the midpoint of $\overline{FG}$ (Given)
2. $\overline{EF}\cong\overline{EG}$ (The midpoint of a line divides it into 2 congruent lines)
3. $\overline{DF}\cong\overline{DG}$ (Given)
4. $\overline{DE}\cong\overline{DE}$ (Identity)
So now we have all 3 sides of $\triangle FED$ are congruent with 3 corresponding sides of $\triangle GED$. Therefore,
5. $\triangle FED\cong\triangle GED$ (SSS)
6. $\angle DEF\cong\angle DEG$ (CPCTC)
However, we see that $\angle DEF+\angle DEG=\angle FEG=180^o$ (since $\overline{FG}$ is a line)
Therefore, the value of each angle must be $90^o$. So,
7. $\angle DEF$ and $\angle DEG$ are both right $\angle$s.
8. $\overline{DE}\bot\overline{FG}$ (if a line intersects another one and creates 2 right angles, then those 2 lines are perpendicular with each other)
