Question

find $m\angle bfc$.
$m\angle bfc = \square^\circ$

Explanation:

Step1: Identify right angle and vertical angles

From the diagram, \( \angle AFG \) is a right angle (\( 90^\circ \)), and \( \angle EFD = 28^\circ \). Also, \( \angle BFC \) and \( \angle EFD \) are vertical angles? Wait, no, actually, since \( AF \perp GC \), \( \angle AFC = 90^\circ \)? Wait, no, let's look again. The line \( AF \) is perpendicular to \( GC \), so \( \angle AFG = \angle AFC = 90^\circ \). The angle between \( EF \) and \( FD \) is \( 28^\circ \), and \( \angle BFC \) and \( \angle EFD \) are vertical angles? Wait, no, \( AB \) and \( ED \) are intersecting lines, so \( \angle BFC \) and \( \angle EFD \) are vertical angles? Wait, no, actually, \( \angle BFC \) and \( \angle EFD \) are vertical angles? Wait, no, let's see: \( AF \) is perpendicular to \( GC \), so \( \angle AFG = 90^\circ \). The angle between \( BF \) and \( CF \): since \( AF \) is perpendicular to \( GC \), \( \angle AFC = 90^\circ \), and the angle between \( EF \) and \( FD \) is \( 28^\circ \), so \( \angle BFC \) should be equal to \( 90^\circ - 28^\circ \)? Wait, no, maybe I got it wrong. Wait, \( AF \) is vertical, \( GC \) is horizontal, so they are perpendicular. The line \( ED \) makes a \( 28^\circ \) angle with \( FD \) (which is vertical down). Then \( AB \) is a line intersecting \( ED \) at \( F \), so \( \angle BFC \) and \( \angle EFD \) are vertical angles? Wait, no, \( \angle BFC \) and \( \angle EFD \): let's check the angles. Since \( AF \perp GC \), \( \angle AFC = 90^\circ \). The angle between \( BF \) and \( AF \): wait, maybe \( \angle BFA \) is equal to \( \angle EFD = 28^\circ \) (vertical angles). Then \( \angle BFC = \angle AFC - \angle BFA = 90^\circ - 28^\circ = 62^\circ \)? Wait, no, \( \angle AFC \) is \( 90^\circ \), \( \angle BFA \) is \( 28^\circ \), so \( \angle BFC = 90^\circ - 28^\circ = 62^\circ \)? Wait, no, let's re-express. The line \( AF \) is perpendicular to \( GC \), so \( \angle AFG = \angle AFC = 90^\circ \). The angle between \( EF \) and \( FD \) is \( 28^\circ \), and \( AB \) and \( ED \) are straight lines intersecting at \( F \), so \( \angle BFC \) and \( \angle EFD \) are vertical angles? Wait, no, \( \angle BFC \) and \( \angle EFD \): if \( AB \) and \( ED \) intersect at \( F \), then \( \angle BFC = \angle EFD \)? But \( \angle EFD \) is \( 28^\circ \)? No, that can't be. Wait, maybe I messed up the diagram. Let's see: \( AF \) is up, \( FD \) is down (vertical), \( GC \) is left-right (horizontal). \( ED \) is a line going from \( E \) (down-left) to \( D \) (down-right? No, \( E \) is down-left, \( D \) is down-right? No, \( E \) is a point on the line going down-left, \( D \) is down-right? Wait, no, the diagram shows \( E \) below \( F \) to the left, \( D \) below \( F \) to the right, so \( ED \) is a line going from \( E \) (left-down) to \( D \) (right-down), making a \( 28^\circ \) angle with \( FD \) (which is vertical down). So \( \angle EFD = 28^\circ \), and \( AB \) is a line going from \( A \) (up) to \( B \) (up-right), intersecting \( ED \) at \( F \). Then \( \angle BFC \) is the angle between \( BF \) (up-right) and \( CF \) (right). Since \( AF \) is up (vertical), \( GC \) is right (horizontal), so they are perpendicular. So \( \angle AFC = 90^\circ \). The angle between \( BF \) and \( AF \) is equal to \( \angle EFD = 28^\circ \) (vertical angles), so \( \angle BFA = 28^\circ \). Then \( \angle BFC = \angle AFC - \angle BFA = 90^\circ - 28^\circ = 62^\circ \). Wait, that makes sense. So \( m\angle BFC = 62^\circ \).

Step2: Calculate \( m\angle BFC \)

Since \( AF \perp GC \), \( \angle AFC = 90^\circ \). \( \angle BFA \) and \( \angle EFD \) are vertical angles, so \( \angle BFA = \angle EFD = 28^\circ \). Then \( m\angle BFC = \angle AFC - \angle BFA = 90^\circ - 28^\circ = 62^\circ \).

Answer:

\( 62 \)