Question

1. the pipe in figure 23 - 6 has been bent so that ( overleftrightarrow{ab}paralleloverleftrightarrow{cd} ). if ( angle abc = 120^{circ} ), what is the measure of ( angle bcd )? explain your answer.

Explanation:

Step1: Draw a line through point \( C \) parallel to \( AB \) and \( CD \)

Let \( CE\parallel AB\), since \( AB\parallel CD \), then \( CE\parallel CD \)

Step2: Use the property of parallel lines and angles

When \( CE\parallel AB \), \(\angle ABC+\angle BCE = 180^{\circ}\) (consecutive - interior angles). Given \(\angle ABC = 120^{\circ}\), then \(\angle BCE=180^{\circ}-\angle ABC = 180 - 120=60^{\circ}\)
When \( CE\parallel CD \), \(\angle ECD = 0^{\circ}\) (because \( CE\) and \( CD\) are parallel and on the same line - like a non - deviated part for the parallel relationship in terms of the angle formed here, actually, we can also use the formula for the sum of interior angles on the same side of two parallel lines. The sum of \(\angle ABC\) and \(\angle BCD\) for the two parallel lines \(AB\) and \(CD\) with the transversal \(BC\) (by constructing the parallel line \(CE\) we can also think of it as using the property that the sum of interior angles on the same side of two parallel lines \(AB\) and \(CD\) is \(180^{\circ}\) if we consider the "broken - line" transversal. Another way: \(\angle BCD=180^{\circ}-\angle ABC\))

Answer:

\(60^{\circ}\)