In the diagram above, angles GAF, FBE, and ECD are all equal in measur

Question

https://gmatclub.com/forum/download/file.php?id=24946 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 Untitled.png

VladimirKarpov · Accepted Answer

Let "O" be the center of the circle
Angle AOG=180

Arcs GF=FE=ED=DC=CD=BA
Total: 6 arcs
180/6=30

So angles GOF=FOE=EOD=DOC=COB=BOA=30

Inscribed angles measure half of their corresponding central angles = 30/2 = 15

anceer · Answer

The OE 
Can any one explain it..

A. This problem relies heavily on the concepts of inscribed and central angles. If all the inscribed angles in the figure (GAF, FBE, ECD, and their "opposite" angles across the circle) are the same measurement, then there are 6 identical inscribed angles in the semicircle. Now, consider if each arclength - AB, BC, CD, etc. - were created by central angles. That would be six 30-degree angles from the center of the circle, and since inscribed angles measure half of their corresponding central angles, that means that the inscribed angles are 15 degrees, making A the correct answer.

Shree9975 · Answer

Hi,

Can any1 please explain why are all arcs equal??
And why are all angles equal??

anceer · Answer

All 3 lines are parallel, and the internal angel FAG EBF and DCE are equal.
so the other 3 angle must be identical
(just for clear understanding take the section ABFG.
you know the distance between AF and BG are equal as BF // AG
so the angle made by BG on AG must be equal to the angle made by FA on AG)
When angles are same the arc must be same...

hope you got it

vipulgoel · Answer

OK now what, this is how you got AB = FG , and BC = EF , but how AB=BC = CD possible??

jazyb · Answer

because these lines are parallel to each other.

attari92 · Answer

as per the given question stem each of the six arch would make equal angle at the circumference of the circle that implies the archs are of equal length. Angle each of them formed at the center must also be equal, which implies each formed an angle of 180/6 degrees and angle formed at the center is double the angle formed at the circumference, and importantly, required angle is at the circumference => angle CFB =180 /6*2= 15 degress -> option A

ShashankDave · Answer

Bunuel VeritasPrepKarishma please confirm, is D placed exactly in the middle of the semicircle? If yes then I have a solution. But before I discuss my solution that took me like years to figure out, please share your explanations and related concepts. Thanks a lot

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION:

A. This problem relies heavily on the concepts of inscribed and central angles. If all the inscribed angles in the figure (GAF, FBE, ECD, and their "opposite" angles across the circle) are the same measurement, then there are 6 identical inscribed angles in the semicircle. Now, consider if each arclength - AB, BC, CD, etc. - were created by central angles. That would be six 30-degree angles from the center of the circle, and since inscribed angles measure half of their corresponding central angles, that means that the inscribed angles are 15 degrees, making A the correct answer.

ScottTargetTestPrep · Answer

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

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In the diagram above, angles GAF, FBE, and ECD are all equal in measur

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In the diagram above, angles GAF, FBE, and ECD are all equal in measur

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