Question

question 9(multiple choice worth 1 points)
(07.01 lc)
for circle b, bg = be, bg is perpendicular to dc, and be is perpendicular to fa. what conclusion can be made?

Explanation:

Step1: Recall the perpendicular - chord bisector theorem

If a line from the center of a circle is perpendicular to a chord, then it bisects the chord. Since \(BG = BE\) (radii of the same circle \(B\)), \(BG\perp DC\), and \(BE\perp FA\).

Step2: Apply the theorem

By the perpendicular - chord bisector theorem, if a radius is perpendicular to a chord, the chords that are equidistant from the center (in this case, since \(BG = BE\)) are congruent.

Answer:

\(\overline{CD}\cong\overline{EB}\) is incorrect. \(\overline{GB}\parallel\overline{EB}\) is incorrect (they are the same line in a sense, but not parallel as parallel lines are distinct). \(\overline{DC}\parallel\overline{FA}\) is not necessarily true from the given information. The correct conclusion is \(\overline{DC}\cong\overline{FA}\) (because chords equidistant from the center of a circle are congruent). So the answer is \(\boldsymbol{\overline{DC}\cong\overline{FA}}\) (the first option).