Question

ab||ef and de||bc. determine ∠cqf.
determine ∠cdb, if ∠acb is 41°.

Explanation:

Step1: Identify Alternate Interior Angles (AB∥EF, DE∥BC)

Since \( AB \parallel EF \) and \( DE \parallel BC \), quadrilateral \( PQBE \) is a parallelogram (opposite sides parallel). So, \( \angle EPQ = \angle PBQ \) (alternate interior angles), but more directly, \( \angle AP E = 39^\circ \), and since \( DE \parallel BC \), \( \angle B = \angle E \) (alternate interior angles for \( DE \parallel BC \) and transversal \( EB \)). Also, \( AB \parallel EF \), so \( \angle BPQ + \angle PQF = 180^\circ \) (consecutive interior angles), but we need \( \angle CQF \).

Wait, better approach: Since \( DE \parallel BC \), \( \angle DP A = \angle B \) (corresponding angles, \( AB \) is transversal). Then, since \( AB \parallel EF \), \( \angle B + \angle BQF = 180^\circ \) (consecutive interior angles). But \( \angle CQF \) is supplementary to \( \angle EQC \), and \( \angle EQC = \angle DP A = 39^\circ \) (vertical angles or corresponding angles). Wait, no: \( \angle DP A = 39^\circ \), and \( DE \parallel BC \), so \( \angle DP A = \angle B \) (corresponding angles). Then \( AB \parallel EF \), so \( \angle B + \angle BQF = 180^\circ \), but \( \angle CQF \) is vertical to \( \angle DQE \), or wait, \( \angle CQF \) and \( \angle DQP \)? No, let's look at the lines: \( DE \) and \( BC \) are parallel, \( AB \) and \( EF \) are parallel. So the angle at \( P \) is \( 39^\circ \), so the angle at \( Q \) ( \( \angle CQF \)) should be \( 180^\circ - 39^\circ = 141^\circ \)? Wait, no: \( \angle AP E = 39^\circ \), so \( \angle DP B = 39^\circ \) (vertical angles). Then, since \( DE \parallel BC \), \( \angle DP B = \angle B \) (alternate interior angles). Then, since \( AB \parallel EF \), \( \angle B + \angle BQF = 180^\circ \) (consecutive interior angles). But \( \angle CQF = \angle BQF \)? No, \( \angle CQF \) is adjacent to \( \angle BQF \)? Wait, maybe I messed up. Let's use linear pairs and parallel lines:

\( AB \parallel EF \), so \( \angle BPQ + \angle PQF = 180^\circ \). \( \angle BPQ = 180^\circ - 39^\circ = 141^\circ \) (since \( \angle AP E = 39^\circ \), linear pair with \( \angle BPQ \)). Then, since \( DE \parallel BC \), \( \angle BPQ = \angle CQF \) (corresponding angles, because \( DE \parallel BC \) and transversal \( PQ \)). Wait, yes! \( DE \parallel BC \), so \( \angle BPQ \) and \( \angle CQF \) are corresponding angles (transversal \( PQ \)). So \( \angle BPQ = 180^\circ - 39^\circ = 141^\circ \), so \( \angle CQF = 141^\circ \).

Wait, let's re-express:

1. \( \angle AP E = 39^\circ \) (given).
2. \( \angle AP E \) and \( \angle BP D \) are vertical angles? No, \( \angle AP E \) and \( \angle DP B \) are vertical angles? Wait, \( A---P---B \) is a straight line, \( D---P---E \) is a straight line? No, the diagram: \( A \) to \( P \) to \( B \) is horizontal, \( D \) to \( P \) to \( E \) is a diagonal? Wait, maybe the correct way is:

Since \( AB \parallel EF \) and \( DE \parallel BC \), the figure is a parallelogram, so \( \angle at P \) (39°) and \( \angle at Q \) ( \( \angle CQF \)) are supplementary? No, wait, \( \angle AP E = 39^\circ \), so \( \angle EP Q = 180^\circ - 39^\circ = 141^\circ \) (linear pair). Then, since \( DE \parallel BC \), \( \angle EP Q = \angle CQF \) (corresponding angles, transversal \( EF \) or \( AB \)). Yes! So \( \angle CQF = 180^\circ - 39^\circ = 141^\circ \).

Step1: Find linear pair of \( 39^\circ \)

\( \angle AP E = 39^\circ \), so \( \angle EP B = 180^\circ - 39^\circ = 141^\circ \) (linear pair, \( A-P-B \) is straight line).

Step2: Corresponding angles ( \( DE \parallel BC \))

Since \( DE \parallel BC \) and \( EF \) (or \( AB \)) is transversal, \( \angle EP B = \angle CQF \) (corresponding angles). Thus, \( \angle CQF = 141^\circ \).

Answer:

\( \angle CQF = \boxed{141^\circ} \)