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For circle A, AF = AG, AF is perpendicular to CD, and AG is perpendicular to EB. What conclusion can be made?

a circle with center A and chords CD and EB, a segment from A to chord CD intersects chord CD at F, and a segment from A to chord EB intersects chord EB at G

segment CD is parallel to segment EB
segment CD is congruent to segment EB
segment AF is perpendicular to segment AG
segment CF is congruent to segment AF

1 answer

Given the information provided:

  1. \( AF = AG \): This means that the distances from the center \( A \) to the points \( F \) and \( G \) on the respective chords \( CD \) and \( EB \) are equal.
  2. \( AF \) is perpendicular to \( CD \): This implies that \( F \) is the midpoint of the chord \( CD \) because the perpendicular from the center of a circle to a chord bisects that chord.
  3. \( AG \) is perpendicular to \( EB \): This implies that \( G \) is the midpoint of the chord \( EB \) for the same reason.

Because \( AF = AG \) and \( F \) and \( G \) are midpoints of the chords \( CD \) and \( EB \) respectively, we can conclude that the segments \( CD \) and \( EB \) must be equidistant from the center \( A \).

From these conditions, we conclude that:

  • The perpendicular distances from the center \( A \) to the two chords \( CD \) and \( EB \) are equal.
  • Since both chords are bisected at points \( F \) and \( G \) and are equidistant from the center, we can deduce that \( CD \) and \( EB \) must also be parallel to each other because they are both at the same distance from the center.

Thus, the best conclusion we can draw is:

The segment CD is parallel to segment EB.