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In the diagram above, angles GAF, FBE, and ECD are all equal in measur [#permalink]
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04 Dec 2014, 05:44
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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 Attachment:
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Last edited by Bunuel on 31 Dec 2017, 14:13, edited 2 times in total.
Edited the question.



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Re: In the diagram above, angles GAF, FBE, and ECD are all equal in measur [#permalink]
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05 Dec 2014, 05:59
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 30degree 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.



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Re: In the diagram above, angles GAF, FBE, and ECD are all equal in measur [#permalink]
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05 Dec 2014, 07:42
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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



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Re: In the diagram above, angles GAF, FBE, and ECD are all equal in measur [#permalink]
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22 Dec 2014, 04:58
Hi,
Can any1 please explain why are all arcs equal?? And why are all angles equal??



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Re: In the diagram above, angles GAF, FBE, and ECD are all equal in measur [#permalink]
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29 Dec 2014, 20:38
Shree9975 wrote: Hi,
Can any1 please explain why are all arcs equal?? And why are all angles equal?? 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



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Re: In the diagram above, angles GAF, FBE, and ECD are all equal in measur [#permalink]
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07 Jun 2015, 22:22
anceer wrote: Shree9975 wrote: Hi,
Can any1 please explain why are all arcs equal?? And why are all angles equal?? 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 OK now what, this is how you got AB = FG , and BC = EF , but how AB=BC = CD possible??



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Re: In the diagram above, angles GAF, FBE, and ECD are all equal in measur [#permalink]
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05 Mar 2016, 19:17
because these lines are parallel to each other. vipulgoel wrote: anceer wrote: Shree9975 wrote: Hi,
Can any1 please explain why are all arcs equal?? And why are all angles equal?? 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 OK now what, this is how you got AB = FG , and BC = EF , but how AB=BC = CD possible??



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Re: In the diagram above, angles GAF, FBE, and ECD are all equal in measur [#permalink]
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23 May 2017, 05:13
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



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Re: In the diagram above, angles GAF, FBE, and ECD are all equal in measur [#permalink]
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17 Nov 2017, 06:55
anceer wrote: Attachment: Untitled.png 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 BunuelVeritasPrepKarishma 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
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Re: In the diagram above, angles GAF, FBE, and ECD are all equal in measur [#permalink]
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31 Dec 2017, 14:13
anceer wrote: 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 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 30degree 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.
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Re: In the diagram above, angles GAF, FBE, and ECD are all equal in measur
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