Question

in the figure below, \\( \overline{ac} \parallel \overline{dg} \\), \\( m\angle cbf = 60^\circ \\), and \\( m\angle def = 140^\circ \\).

what is the \\( m\angle bfe \\), indicated by \\( \theta \\)?

select one answer
a \\( 60^\circ \\)
b \\( 100^\circ \\)
c \\( 120^\circ \\)
d \\( 140^\circ \\)

note: figure not drawn to scale

Explanation:

Step1: Draw a line parallel to \( \overline{AC} \) and \( \overline{DG} \) through point \( F \), say \( \overline{FH} \), so \( \overline{AC} \parallel \overline{FH} \parallel \overline{DG} \).

Step2: For \( \overline{AC} \parallel \overline{FH} \), \( \angle CBF \) and \( \angle BFH \) are alternate interior angles, so \( m\angle BFH = m\angle CBF = 60^\circ \).

Step3: For \( \overline{FH} \parallel \overline{DG} \), \( \angle DEF \) and \( \angle HFE \) are same - side interior angles. Since \( m\angle DEF = 140^\circ \), \( m\angle HFE=180^\circ - 140^\circ = 40^\circ \)? Wait, no, actually, \( \angle DEF \) and \( \angle HFE \) should be supplementary? Wait, no, let's correct. \( \overline{DG} \) is a straight line, \( \angle DEG = 180^\circ \), \( m\angle DEF = 140^\circ \), so \( m\angle FEG=180 - 140 = 40^\circ \)? No, better to use the fact that when we have two parallel lines and a transversal, consecutive interior angles are supplementary. Wait, actually, the sum of the angles around the "zig - zag" path. The sum of the interior angles of a polygon? Wait, another approach: The sum of the angles in the figure. The total turn from \( \overline{AC} \) to \( \overline{DG} \) should be considered. The angle at \( B \) is \( 60^\circ \) (upward), the angle at \( E \) is \( 140^\circ \) (but we need the angle adjacent to \( \angle DEF \) which is \( 180 - 140 = 40^\circ \) downward? Wait, no. Let's use the formula for the sum of angles in a "trapezoid - like" figure with two parallel sides. The sum of the angles \( \angle CBF + \angle BFE+ (180^\circ - \angle DEF)= 360^\circ - 180^\circ\)? Wait, no. Let's use the method of drawing a parallel line.

Let \( \overline{FK} \parallel \overline{AC} \parallel \overline{DG} \) through \( F \). Then \( \angle CBF \) and \( \angle BFK \) are alternate interior angles, so \( \angle BFK = 60^\circ \). And \( \angle KFE \) and \( \angle DEF \) are same - side interior angles? No, \( \angle DEF \) and \( \angle KFE \) should be supplementary because \( \overline{FK}\parallel\overline{DG} \). Wait, \( m\angle DEF = 140^\circ \), so \( m\angle KFE = 180 - 140=40^\circ \)? No, that can't be. Wait, I made a mistake. The correct way: The sum of the angles \( \angle CBF + \angle BFE+(180 - \angle DEF)= 180\times2 \)? Wait, no. Let's think of the figure as a transversal cutting two parallel lines, and the angle \( \theta \) is equal to \( 60^\circ+(180 - 140)^\circ=60 + 40=100^\circ \). Wait, let's recast:

The sum of the interior angles for the path \( B - F - E \) with the two parallel lines \( AC \) and \( DG \). The formula for the angle in such a "zig - zag" (a type of polygon - like figure between two parallel lines) is that the sum of the angles on one side is equal to the sum of the angles on the other side. Wait, actually, when you have two parallel lines \( AC \) and \( DG \), and the segments \( BC \), \( BF \), \( FE \), \( EG \), the measure of \( \angle BFE \) is equal to \( 180^\circ-(180^\circ - 60^\circ)-(180^\circ - 140^\circ) \)? No, better to use the fact that the sum of the angles \( \angle CBF + \angle BFE+(180^\circ - \angle DEF)= 180^\circ\times2 \)? No, let's use the parallel line method correctly.

Draw a line \( \overline{FL} \parallel \overline{AC} \) (and hence \( \overline{DG} \)) through \( F \). Then \( \angle CBF=\angle BFL = 60^\circ \) (alternate interior angles). And \( \angle LFE = 180^\circ - \angle DEF \)? Wait, \( \angle DEF = 140^\circ \), so \( \angle LFE=180 - 140 = 40^\circ \)? No, that's not right. Wait, \( \overline{DG} \) is a straight line, so \( \angle DEG = 180^\circ \), \( \angle DEF = 140^\circ \), so \( \angle FEG = 40^\circ \). Since \( \overline{FL}\parallel\overline{DG} \), \( \angle LFE=\angle FEG = 40^\circ \)? No, that's not. Wait, I think I messed up the direction. Let's start over.

The sum of the angles around the points: The total angle for a "polygon" with vertices \( B \), \( F \), \( E \) and the two parallel lines. The formula for the angle between two parallel lines with a transversal - like path: \( \angle BFE= \angle CBF+(180^\circ - \angle DEF) \). So \( \angle CBF = 60^\circ \), \( 180-\angle DEF = 180 - 140 = 40^\circ \), so \( \angle BFE=60 + 40=100^\circ \).

Answer:

B. \( 100^\circ \)