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

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Accepted Answer

# Explanation:
## Step1: Identify Alternate Interior Angles
Since $ AB \parallel EF $ and $ DE \parallel BC $, quadrilateral $ PQBE $ (or the parallelogram formed) has $ \angle AP E = 39^\circ $. First, use the property of parallel lines $ AB \parallel EF $ and transversal $ DE $ (or the other lines) to find related angles. Wait, actually, since $ DE \parallel BC $, the angle at $ P $ ( $ \angle AP E = 39^\circ $) and the angle at $ E $ would be equal (alternate interior angles), but maybe better to look at the transversal for $ \angle CQF $.

Wait, let's correct. Let's consider $ AB \parallel EF $ and $ DE $ as a transversal, but also $ DE \parallel BC $. Wait, actually, the key is that $ \angle AP P' $ (wait, the angle given is $ 39^\circ $ at $ P $ between $ AB $ and the line to $ E $). Then, since $ DE \parallel BC $, the angle at $ P $ ( $ 39^\circ $) and the angle at $ Q $ (related to $ \angle CQF $): Wait, $ \angle CQF $ is a linear pair or supplementary? Wait, no. Let's think again.

Since $ AB \parallel EF $, and $ DE $ is a transversal, but maybe the quadrilateral is a parallelogram, so $ \angle B = 39^\circ $? No, wait, the angle at $ P $ is $ 39^\circ $, so the angle inside the parallelogram at $ B $ would be equal (since $ DE \parallel BC $, alternate interior angles). Then, for $ \angle CQF $, since it's a straight line with the angle adjacent to it, but actually, $ \angle CQF $ and the angle at $ Q $ (inside the parallelogram) are supplementary? Wait, no. Wait, the correct approach:

1. Since $ AB \parallel EF $ and $ DE \parallel BC $, the figure $ PBQE $ is a parallelogram (opposite sides parallel). So $ \angle BP E = 180^\circ - 39^\circ = 141^\circ $? No, wait, $ \angle AP E = 39^\circ $, so $ \angle BP E = 180^\circ - 39^\circ = 141^\circ $? No, $ AB $ is a straight line, so $ \angle AP E + \angle BP E = 180^\circ $, so $ \angle BP E = 180 - 39 = 141^\circ $. But since $ DE \parallel BC $, $ \angle BP E = \angle CQF $ (corresponding angles)? Wait, no, $ DE \parallel BC $, so the transversal is $ EF $? Wait, maybe I messed up.

Wait, the correct way: $ \angle CQF $ is supplementary to the angle that is equal to $ 39^\circ $? No, wait, let's look at the diagram. The angle at $ P $ is $ 39^\circ $, between $ AB $ and the line to $ E $. Then, since $ DE \parallel BC $, the angle at $ P $ ( $ 39^\circ $) and the angle at $ Q $ (the one adjacent to $ \angle CQF $): Wait, $ \angle CQF $ is a straight angle with the angle inside the parallelogram. Wait, actually, $ \angle CQF $ and the angle that is equal to $ 180^\circ - 39^\circ $? No, wait, no. Let's recall that in parallel lines, consecutive interior angles are supplementary, alternate interior are equal, corresponding are equal.

Wait, let's start over. Given $ AB \parallel EF $ and $ DE \parallel BC $. The angle at $ P $ ( $ \angle AP E = 39^\circ $) is an alternate interior angle with the angle at $ E $ (since $ AB \parallel EF $ and $ DE $ is transversal). Then, since $ DE \parallel BC $, the angle at $ E $ is equal to the angle at $ B $ (alternate interior angles). Then, for $ \angle CQF $, since $ BC \parallel DE $, and $ EF $ is a transversal, $ \angle CQF $ and the angle at $ B $ (which is $ 39^\circ $): Wait, no, $ \angle CQF $ is a linear pair with the angle that is equal to $ 39^\circ $? Wait, no, I think I made a mistake. Let's look at the diagram again (as per the user's image):

- $ AB $ is horizontal left (A to B), $ EF $ is horizontal right (E to F).
- $ DE $ is a line from D (top) to E (bottom left), passing through P (on AB) and Q (on EF).
- $ BC $ is a line from B (right) to C (bottom left), parallel to DE.
- The angle at P between AB and the line to E is $ 39^\circ $.

So, $ AB \parallel EF $, transversal is DE (or the line through P and Q). So $ \angle AP E = 39^\circ $, so $ \angle EP Q = 39^\circ $ (vertical angles? No, $ \angle AP E $ and $ \angle BP Q $ are vertical? Wait, no, $ \angle AP E $ is at P, between A (left) and E (down to E). Then, since $ DE \parallel BC $, $ \angle BP Q = \angle BC Q $ (alternate interior angles), but maybe better to find $ \angle CQF $.

$ \angle CQF $ is at Q, between C (down left) and F (right). So the line CQ is part of DE (or BC? Wait, DE is parallel to BC, so CQ is part of BC? No, DE is from D to E, BC is from B to C, so DE and BC are parallel. So the transversal is EF (for DE and BC). So $ \angle CQF $ and the angle at B ( $ \angle B $): Wait, $ AB \parallel EF $, so $ \angle B = \angle EQF $ (corresponding angles), but $ \angle EQF $ and $ \angle CQF $ are supplementary (linear pair). Wait, no, $ \angle EQF $ and $ \angle CQF $ are adjacent and form a linear pair, so they sum to $ 180^\circ $. But $ \angle B $ is equal to $ 39^\circ $ (since $ DE \parallel BC $, alternate interior angles with $ \angle AP E = 39^\circ $). Wait, $ \angle AP E = 39^\circ $, and since $ AB \parallel EF $, $ \angle E = \angle AP E = 39^\circ $ (alternate interior angles). Then, since $ DE \parallel BC $, $ \angle B = \angle E = 39^\circ $ (alternate interior angles). Then, $ AB \parallel EF $, so $ \angle B = \angle EQF = 39^\circ $ (corresponding angles). Then, $ \angle CQF = 180^\circ - \angle EQF = 180^\circ - 39^\circ = 141^\circ $? Wait, no, that can't be. Wait, maybe I got the direction wrong.

Wait, no, let's use the correct theorem: When two parallel lines are cut by a transversal, consecutive interior angles are supplementary. Wait, $ AB \parallel EF $, transversal is the line through P and Q (DE). So $ \angle AP E $ and $ \angle EP F $ (wait, no). Wait, the angle at P is $ 39^\circ $, so the angle inside the parallelogram (between AB and BC) is $ 39^\circ $, then since $ AB \parallel EF $, the angle at F ( $ \angle CQF $) would be supplementary to that? Wait, no, let's do it step by step.

Step 1: Find $ \angle BP E $. Since $ \angle AP E = 39^\circ $ and $ AB $ is a straight line, $ \angle BP E = 180^\circ - 39^\circ = 141^\circ $.

Step 2: Since $ DE \parallel BC $, $ \angle BP E = \angle CQF $ (corresponding angles, because AB is parallel to EF, and DE/BC are parallel, so the angle at P ( $ \angle BP E $) and the angle at Q ( $ \angle CQF $) are corresponding angles). Wait, yes! Because $ AB \parallel EF $, and the two parallel lines DE and BC are cut by the transversal AB (or EF). Wait, no, the transversal for DE and BC would be AB or EF. Since $ AB \parallel EF $, and DE || BC, then $ \angle BP E $ (at P, between AB and DE) and $ \angle CQF $ (at Q, between EF and BC) are corresponding angles, so they are equal. Therefore, $ \angle CQF = \angle BP E = 141^\circ $.

Wait, that makes sense. So:

## Step1: Calculate $ \angle BP E $
$ \angle AP E = 39^\circ $, and $ \angle AP E + \angle BP E = 180^\circ $ (linear pair), so:
$ \angle BP E = 180^\circ - 39^\circ = 141^\circ $

## Step2: Use Corresponding Angles (DE || BC, AB || EF)
Since $ DE \parallel BC $ and $ AB \parallel EF $, $ \angle CQF $ and $ \angle BP E $ are corresponding angles, so:
$ \angle CQF = \angle BP E = 141^\circ $

# Answer:
$ \boldsymbol{141^\circ} $

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Explanation:

Step1: Identify Alternate Interior Angles

Answer: