Question

question
in the diagram below, \\(\overline{ac} \parallel \overline{dg}\\), \\(\overline{be} \cong \overline{bf}\\) and \\(\text{m}\angle ebf = 50^\circ\\). find \\(\text{m}\angle hbc\\).

step | angle | reason
1 | \\(\text{m}\angle ebf = 50^\circ\\) | given
try | \\(\text{m}\angle \square = \square^\circ\\) | select a reason

diagram: lines ac and dg are parallel, be is congruent to bf, angle ebf is 50 degrees. points c, b, a are colinear; g, f, e, d are colinear; h is a point above b, i is a point below f. the figure has angle ebf labeled 50 degrees.

you may assume lines that appear straight are straight, but the figure is not otherwise drawn to scale.

Explanation:

Step1: Analyze triangle BEF

Since \(\overline{BE} \cong \overline{BF}\), \(\triangle BEF\) is isosceles with \(\angle BEF=\angle BFE\). The sum of angles in a triangle is \(180^\circ\), so \(\angle BEF + \angle BFE+\angle EBF = 180^\circ\). Given \(\angle EBF = 50^\circ\), then \(2\angle BEF=180^\circ - 50^\circ=130^\circ\), so \(\angle BEF = 65^\circ\).

Step2: Use parallel lines (alternate interior angles)

Because \(\overline{AC} \parallel \overline{DG}\), \(\angle HBC=\angle BEF\) (alternate interior angles). So \(m\angle HBC = 65^\circ\).

Answer:

\(m\angle HBC = \boldsymbol{65^\circ}\)