## Elementary Geometry for College Students (6th Edition)

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 $\overline{DE}\bot\overline{FG}$
*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)