Chapter 5.2, Problem 12CE

To determine

To explain: $\overleftrightarrow{HG}$ and $\overleftrightarrow{EF}$ are always parallel.

Explanation of Solution

Given information:

To draw parallel lines, are constructed so that $EF=HG$ and $HE=GF$ . Since there are hinges at points $E,F,G,H$ , can vary at distance between $\overleftrightarrow{HG}$ and $\overleftrightarrow{EF}$ .

Given figure,

Calculation:

In a quadrilateral $HEFG$ ,

$\begin{array}{l}HG=EF\\ HE=GF\end{array}$ ....(1)

It is required to show that $HG\parallel EF$ .

From equation (1) one can conclude that in a quadrilateral $HEFG$ , the opposite sides are congruent.

Note that in a quadrilateral, opposite sides are congruent then that quadrilateral is a parallelogram.

So it follows that given quadrilateral $HEFG$ is a parallelogram.

In a parallelogram opposite side are parallel.

So it implies that,

  $\begin{array}{l}HG\parallel EF\\ HE\parallel GF\end{array}$

Hence, it is proved that $HG\parallel EF$ .