AAPL wrote:Veritas Prep



In the diagram above, angles GAF, FBE, and ECD are all equal in measure. If line AG is the diameter of the circle and parallel to lines BF and CE, what is the measure of angle CFB?

A. 15
B. 20
C. 25
D. 30
E. 35

OA A

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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AAPL wrote:Veritas Prep



In the diagram above, angles \(GAF, FBE, and ECD are all equal in measure. If line AG is the diameter of the circle and parallel to lines BF and CE, what is the measure of angle CFB?

A. 15
B. 20
C. 25
D. 30
E. 35

OA A

We see that angles GAF, FBE, and ECD are inscribed angles, and their respective intercepted arcs are FG, EF and DE. Since angle GAF = angle FBE = angle ECD, arc FG = arc EF = arc DE. We see that arc DG is a quarter circle, which is 90 degrees; therefore, arc FG = arc EF = arc DE = 30 degrees. Since an inscribed angle is half its intercepted arc, angle GAF = angle FBE = angle ECD = 15 degrees. Since angle CFB = angle FBE, angle CFB = 15 degrees.

Answer: A