The bisector of$\angle EFG$ and$\angle EGF$ meet at $I$.

1. If role="math" localid="1649315963251" $m\angle EFG=40$find $m\angle FIG$.
2. If$m\angle EFG=50$ find $m\angle FIG$.
3. Generalize your results in (a) and (b).
   
   

Consider a $\mathrm{\Delta }EFG$, such that,

width="291" height="92" role="math">m∠EGF+m∠EFG+m∠FEG=180m∠EGF+40+80=180m∠EGF=180−120=60

Angle bisector is a line which divides an angle into two equal parts.

Here, $FI$is the bisector of$\angle EFG$ and$GI$ is the bisector of $\angle EGF$, such that,

$\begin{array}{c}m\angle IGF=\frac{1}{2}m\angle EGF\\ m\angle IFG=\frac{1}{2}m\angle EFG\end{array}$

Consider a $\mathrm{\Delta }FIG$, such that,

$$\begin{gathered} m\angle FIG+m\angle IGF+m\angle IFG=180 \\ m\angle FIG+\frac{1}{2}m\angle EGF+\frac{1}{2}m\angle EFG=180 \\ m\angle FIG+\frac{1}{2}\times 60+\frac{1}{2}\times 40=180 \\ m\angle FIG=180-50 \\ =130 \end{gathered}$$

Therefore,if$m\angle EFG=40$ then $m\angle FIG=130$.

Consider a $\mathrm{\Delta }EFG$, such that,

$$\begin{gathered} m\angle EGF+m\angle EFG+m\angle FEG=180 \\ m\angle EGF+50+80=180 \\ m\angle EGF=180-130 \\ =50 \end{gathered}$$

Angle bisector is a line which divides an angle into two equal parts.

Here, $FI$is the bisector of$\angle EFG$and$GI$ is the bisector of $\angle EGF$, such that,

$\begin{array}{c}m\angle IGF=\frac{1}{2}m\angle EGF\\ m\angle IFG=\frac{1}{2}m\angle EFG\end{array}$

Consider a $\mathrm{\Delta }FIG$, such that,

$$\begin{gathered} m\angle FIG+m\angle IGF+m\angle IFG=180 \\ m\angle FIG+\frac{1}{2}m\angle EGF+\frac{1}{2}m\angle EFG=180 \\ m\angle FIG+\frac{1}{2}\times 50+\frac{1}{2}\times 50=180 \\ m\angle FIG=180-50 \\ =130 \end{gathered}$$

Therefore,if$m\angle EFG=50$ then $m\angle FIG=130$.

Here, $FI$is the bisector of$\angle EFG$ and$GI$ is the bisector of $\angle EGF$, such that,

$\begin{array}{c}m\angle IGF=\frac{1}{2}m\angle EGF\\ m\angle IFG=\frac{1}{2}m\angle EFG\end{array}$

Consider a $\mathrm{\Delta }FIG$, such that,

$$\begin{gathered} m\angle FIG+m\angle IGF+m\angle IFG=180 \\ \begin{array}{c}m\angle FIG+\frac{1}{2}m\angle EGF+\frac{1}{2}m\angle EFG=180\\ m\angle FIG+\frac{1}{2}\left(m\angle EGF+m\angle EFG\right)=180\end{array} \end{gathered}$$

Consider a $\mathrm{\Delta }EFG$, such that,

$$\begin{gathered} m\angle EGF+m\angle EFG+m\angle FEG=180 \\ m\angle EGF+m\angle EFG=180-m\angle FEG \end{gathered}$$

Substitute$180-m\angle FEG$ for$\left(m\angle EGF+m\angle EFG\right)$ in step 2.

$$\begin{gathered} m\angle FIG+\frac{1}{2}\left(180-m\angle FEG\right)=180 \\ m\angle FIG+\frac{1}{2}\left(180-80\right)=180 \\ m\angle FIG=180-50 \\ =130 \end{gathered}$$

Hence, $m\angle FIG$is independent of $m\angle EFG$and is always equal to $130$.