Question

for circle c, cg = ce, cg is perpendicular to fb, and ce is perpendicular to da. what conclusion can be made? fg≅co, gc∥ce, dc∥ca, fb≅da

Explanation:

Step1: Recall circle - related properties

In a circle, if a line from the center is perpendicular to a chord, it bisects the chord.

Step2: Analyze given perpendicular - chord relationships

Since $CG\perp FB$ and $C$ is the center of the circle, $CG$ bisects $FB$. Also, since $CE\perp DA$ and $C$ is the center of the circle, $CE$ bisects $DA$. Given $\overline{FB}\cong\overline{DA}$, we can conclude that the perpendicular distances from the center of the circle to congruent chords are equal. That is, $CG = CE$. Also, because $CG = CE$ and $CG\perp FB$, $CE\perp DA$, and $\overline{FB}\cong\overline{DA}$, the arcs intercepted by these congruent chords are congruent. So, $\overset{\frown}{FB}\cong\overset{\frown}{DA}$.

Step3: Consider parallel - line implications

Since $GC\parallel CE$ and $DC\parallel CA$, and $C$ is the center of the circle, we know that the chords $FB$ and $DA$ are parallel. Because $CG\perp FB$ and $CE\perp DA$ and $GC\parallel CE$, the lines $FB$ and $DA$ are parallel.

Answer:

The chords $\overline{FB}$ and $\overline{DA}$ are parallel, $\overset{\frown}{FB}\cong\overset{\frown}{DA}$, and the perpendicular distances from the center of the circle to $\overline{FB}$ and $\overline{DA}$ are equal ($CG = CE$).