Question

in the diagram below, $overline{ad}paralleloverline{eg}$, $mangle bfc = 75^{circ}$ and $mangle gfc = 38^{circ}$. find $mangle fbc$.
step\tangle\treason
1\t$mangle bfc = 75^{circ}$\tgiven
2\t$mangle gfc = 38^{circ}$\tgiven
try $manglesquare=square^{circ}$\tselect a reason
vertical angles
linear pair (or triple)
congruent angles
sum of angles in a triangle
sum of angles in a quadrilateral
base angle of an isosceles triangle
angle bisector definition
corresponding angles
alternate interior angles
alternate exterior angles
same - side interior angles
angle addition postulate
angle subtraction postulate
perpendicular lines form right angles

Explanation:

Step1: Recall angle - sum property of a triangle

The sum of the interior angles of a triangle is 180°. In \(\triangle BFC\), we know two of the angles and want to find the third. Let's first note the given angles: \(\angle BFC = 75^{\circ}\) and assume we are dealing with \(\triangle BFC\) where we need to find \(\angle FBC\).

Step2: Use the angle - sum formula

In \(\triangle BFC\), if we let \(\angle FBC=x\), \(\angle BFC = 75^{\circ}\), and assume the third - angle (not given in name but we know the sum property), we have \(x+\angle BFC+\angle FCB = 180^{\circ}\). First, we need to find \(\angle FCB\). Since \(\angle GFC\) and \(\angle FCB\) are alternate interior angles (because \(\overline{AD}\parallel\overline{EG}\)), \(\angle FCB=\angle GFC = 38^{\circ}\) (alternate interior angles are congruent when two parallel lines are cut by a transversal).
Now, substituting into the angle - sum formula for \(\triangle BFC\): \(x + 75^{\circ}+38^{\circ}=180^{\circ}\).
We can solve for \(x\):

$$
LATEXBLOCK0
$$

Answer:

\(m\angle FBC = 67^{\circ}\)