Assume that O is the center of the circle and lies at the midpoint of AG. Since angles, GAF, FBE, and ECD are angles formed by arcs GF, FE, and ED and are equal, the arcs GF, FE, and ED must be equal. Thus, the arcs GF, FE, and ED must also make equal angles on the center. Thus, /_GOF + /_FOE + /_EOD = 90º. Now, since /_GOF = /_FOE = /_EOD, each is equal to 30º.
The same goes for other arcs AB, BC and CD. Thus, /_COB = 30º.
The arc BC forms the angle /_COB at the center and forms the angle /_CFB at the periphery; thus, /_CFB = 1/2 of /_COB = 1/2 of 30º = 15º
The correct answer: A
Hope this helps!
-Jay
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