Answer:
segment EF ≅ segment GF because both segments are the same distance from the center of circle A ⇒ answer B
Step-by-step explanation:
* Lets revise some facts in the circle
- The perpendicular segment from the center of the circle to a chord
bisects it
- A segment from the center of a circle to the midpoint of a cord is
perpendicular to it
- Congruent chords in a circle are equidistant from the center of the circle,
that means the perpendicular distances from the center of the circle
to the chords are equal
- If two chords in a circle are equidistant from the center of the circle
<em>( the perpendicular distances from the center of the circle to the </em>
<em> chords are equal) </em>, then they are congruent
* Lets solve the problem
∵ ∠EDA is right angle
∵ AD ⊥ FE
∵ ∠GHA is right angle
∵ AH ⊥ FG
∵ AD = AH
∵ EF and GF are chords in the circle A
∴ The chords EF and GF are equidistant from the center of the circle A
- By using the bold fact above
∴ EF ≅ GF
* segment EF ≅ segment GF because both segments are the same
distance from the center of circle A