91Ó°ÊÓ

Consider a $\mathrm{Δ}EFG$, 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$∠EFG$ and$GI$ is the bisector of $∠EGF$, such that,

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

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

$$\begin{gathered} m∠FIG+m∠IGF+m∠IFG=180 \\ m∠FIG+\frac{1}{2}m∠EGF+\frac{1}{2}m∠EFG=180 \\ m∠FIG+\frac{1}{2}×60+\frac{1}{2}×40=180 \\ m∠FIG=180−50 \\ =130 \end{gathered}$$

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

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

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

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

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

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

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

$$\begin{gathered} m∠FIG+m∠IGF+m∠IFG=180 \\ m∠FIG+\frac{1}{2}m∠EGF+\frac{1}{2}m∠EFG=180 \\ m∠FIG+\frac{1}{2}×50+\frac{1}{2}×50=180 \\ m∠FIG=180−50 \\ =130 \end{gathered}$$

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

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

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

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

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

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

$$\begin{gathered} m∠EGF+m∠EFG+m∠FEG=180 \\ m∠EGF+m∠EFG=180−m∠FEG \end{gathered}$$

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

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

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