Question

a parallel - ruler, shown to the right, can be used to draw parallel lines. the distance between the parallel segments ab and dc can vary. the ruler is constructed so that the distance between a and b equals the distance between d and c. the distance between a and c is the same as the distance between b and d. explain why ab and dc are always parallel.
complete the sentences below.
when the parallel ruler is open at any setting, the distance bc is (square). it is given that ab = (square) and ac = (square), so (\triangle abccong\trianglesquare) by (square). hence, (angle abcconganglesquare) by (square). because these are (square) angles formed by ab and cd with transversal line bc, ab(parallel)dc.

Explanation:

Step1: Identify equal - side conditions

We know that \(AB = DC\) and \(AC = BD\) (by the construction of the parallel - ruler). Also, \(BC\) is common to \(\triangle ABC\) and \(\triangle DCB\).

Step2: Prove triangle congruence

In \(\triangle ABC\) and \(\triangle DCB\), we have \(AB = DC\), \(AC = BD\), and \(BC=CB\). So, \(\triangle ABC\cong\triangle DCB\) by the Side - Side - Side (SSS) congruence criterion.

Step3: Identify equal angles

Since \(\triangle ABC\cong\triangle DCB\), then \(\angle ABC=\angle DCB\) by corresponding parts of congruent triangles are congruent (CPCTC).

Step4: Use angle - relationship for parallel lines

\(\angle ABC\) and \(\angle DCB\) are alternate interior angles formed by lines \(AB\) and \(DC\) with transversal \(BC\). When alternate interior angles are equal, the two lines are parallel. So, \(AB\parallel DC\).

Answer:

When the parallel - ruler is in any setting, the distance \(BC\) is common. It is given that \(AB = DC\) and \(AC = BD\), so \(\triangle ABC\cong\triangle DCB\) by SSS. Hence, \(\angle ABC=\angle DCB\) by CPCTC. Because these are alternate interior angles formed by \(AB\) and \(DC\) with transversal line \(BC\), \(AB\parallel DC\).