Question

if a is the center of the circle, then which statement explains how $overline{ed}$ is related to $overline{fd}$? $overline{ed}=overline{fd}$ because $overline{ef}$ is perpendicular to a radius of circle a. $overline{ed}=overline{fd}$ because $widehat{ef}=widehat{gf}$. $overline{ed}=overline{fd}$ because the inscribed angles that create the segments are congruent. $overline{ed}=overline{fd}$ because the tangents that create the $overline{ef}$ share a common endpoint.

Explanation:

In a circle, if a line from the center of the circle is perpendicular to a chord, it bisects the chord. Here, since \(A\) is the center of the circle and \(AD\perp EF\), \(D\) is the mid - point of \(EF\), so \(\overline{ED}=\overline{FD}\). The correct reason is that a line from the center perpendicular to a chord bisects the chord. Among the given options, the first option is correct as it is based on the property of a circle where a chord perpendicular to a radius (from the center to the chord) is bisected.

Answer:

\(\overline{ED}=\overline{FD}\) because \(\overline{EF}\) is perpendicular to a radius of circle \(A\).