Chapter 5.1, Problem 8CE

To determine

To proof: weather the quad $EFGH$ can be a parallelogram.

Answer to Problem 8CE

The quad $EFGH$ cannot be a parallelogram.

Explanation of Solution

Given information:

System figure:

  

Calculation:

The figure shows a quadrilateral $EFGH$ such that $m\angle HEF={65}^{\circ }$ and $m\angle EHG={115}^{\circ }$ .

Find the sum of the above two angles.

  $\begin{array}{cc}m\angle HEF+m\angle EHG& ={65}^{\circ }+{115}^{\circ }\\ & ={180}^{\circ }\end{array}$

As per Theorem $3.6,$ if two lines are cut by a transversal and same-side interior angles are supplementary, then the lines are parallel.

For quadrilateral EFGH, HG and EF are cut by transversal HE and same-side interior angles are supplementary.

Therefore, $HG\parallel EF$

However, nothing more can be determined about quad $EFGH$ .

Therefore, quad $EFGH$ cannot be a parallelogram.