Chapter 3.4, Problem 28WE

To determine

To Prove: $GK\parallel HI$

Explanation of Solution

Given information:

  $\begin{array}{l}\text{Given:}\overrightarrow{GK}\text{bisects}\angle JG\text{;}\hfill \\ m\angle H=m\angle l\hfill \end{array}$

  

In a triangle the exterior angle equal to the sum of remote interior angles.

  $\begin{array}{l}m\angle JGI=m\angle H+m\angle I\hfill \\ m\angle JGI=m\angle I+m\angle I(m\angle H=m\angle I)\hfill \\ m\angle JGI=2m\angle I\hfill \end{array}$

  $\overrightarrow{GK}$ is a bisector of $\angle JGI$ which means that

  $\begin{array}{cc}m\angle KGI& =\frac{1}{2}\angle JGI\\ & =\frac{1}{2}(2m\angle I)\\ & =m\angle I\\ m\angle KGI& =m\angle I\end{array}$

This means alternative angles are equal.

If two lines are intersected by a transversal line and a pair of alternative angles are equal then the lines are parallel

That is $GK\parallel HI$ .