Question

Parallel rulers, used to draw parallel lines, are constructed so that $\mathrm{EF}=\mathrm{HG}$and $\mathrm{HE}=\mathrm{GF}$. Since there are hinges at points E, F, G and H, you can vary the distance between role="math" localid="1637730770232" $\overleftrightarrow{HG}$and role="math" localid="1637730792914" $\overleftrightarrow{\mathrm{EF}}$. Explain why $\overleftrightarrow{\mathrm{HG}}$and $\overleftrightarrow{\mathrm{EF}}$are always parallel.

Short answer

It is being given that $\mathrm{EF}=\mathrm{HG}$and $\mathrm{HE}=\mathrm{GF}$.

Therefore, it can be said that $\mathrm{EF}\cong \mathrm{HG}$and $\mathrm{HE}\cong \mathrm{GF}$.

Therefore, the opposite sides of the quadrilateral EFGH are congruent.

If both pairs of opposite sides of the quadrilateral are congruent, then the quadrilateral is a parallelogram.

As the opposite sides of the quadrilateral EFGH are congruent, therefore, the quadrilateral EFGH is a parallelogram.

In the parallelogram, the pair of opposite sides are parallel.

As, EFGH is a parallelogram, therefore $\mathrm{EF}\parallel \mathrm{HG}$and $\mathrm{HE}\parallel \mathrm{GF}$.

Therefore, $\mathrm{EF}\parallel \mathrm{HG}$.

Hence proved that $\overleftrightarrow{\mathrm{HG}}$and $\overleftrightarrow{\mathrm{EF}}$are always parallel.

Step-by-step solution

Step 1. Observe the given diagram.

The given diagram is:

Step 2. Description of step.

It is being given that $\mathrm{EF}=HG$and $\mathrm{HE}=GF$.

Therefore, it can be said that $\mathrm{EF}\cong HG$and $\mathrm{HE}\cong GF$.

Therefore, the opposite sides of the quadrilateral EFGH are congruent.

If both pairs of opposite sides of the quadrilateral are congruent, then the quadrilateral is a parallelogram.

As, the opposite sides of the quadrilateral EFGH are congruent, therefore, the quadrilateral EFGH is a parallelogram.

Step 3. Description of step.

In the parallelogram, the pair of opposite sides are parallel.

As, EFGH is a parallelogram, therefore $\mathrm{EF}\parallel HG$and $\mathrm{HE}\parallel GF$.

Therefore, $\mathrm{EF}\parallel HG$.

Hence proved that $\overleftrightarrow{HG}$and $\overleftrightarrow{EF}$are always parallel.