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In the diagram above, AB is parallel to EH, and BD is parall [#permalink]
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08 May 2014, 11:57
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68% (03:00) correct 32% (03:16) wrong based on 183 sessions
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In the diagram above, AB is parallel to EH, and BD is parallel to FH. Also, AB = BC, and EF = FH. If ∠EGC = 70°, then ∠D = (A) 65° (B) 70° (C) 75° (D) 80° (E) 85°For other Geometry practice questions, as well as the OE to this particular question, see: http://magoosh.com/gmat/2014/gmatgeome ... problems/Mike
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Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]
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09 May 2014, 00:09
1. AB//EH ==> ^A = ^EGC = 70. AB = BC ==> ^ACB = 70 = ^DCG 2. ^ABC = 180  (70+70) = 40 3. AB//EH & BD//FH ==> ^EHF = ^ABC = 40 4. EF = FH ==> ^FEH = ^EHF = 40 ==> ^EFH = 180  (40 + 40) = 100 5. BD // FH ==> ^D + ^EFH = 180 ==> ^D = 180  ^EFH = 180  100 = 80
OR
4. When ^FEH = 40 ==> ^DEG = 180  40 = 140 5. ^D = 360  (^DEG + ^EGC + ^GCD) = 360  (140 + 70 + 70) = 360  280 = 80
SO D is indeed the correct answer.



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Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]
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11 May 2014, 09:51
ABEH and AG is transversal => Angle EGA = Angle GAB = 70 degree [ Alternate Angles] Angle GAB = Angle CAB = Angle ACB = 70 (AB = CB) => Angle B = 40 again, ABEH & BDFH => Angle DBA = Angle FHE = 40 degree Therefore, Angle EGH = (180  2*40) = 100 degrees Now, BDFH and DF is transversal, so Angle BDF + Angle HFE= 180 => BDF + 100 = 180 => CDE = 80 Answe : D
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Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]
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14 Aug 2014, 02:03
Answer = 80 Refer diagram below
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Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]
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03 Apr 2016, 11:33
what a nightmare the question is..tried multiple times to draw on my paper..
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Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]
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03 Apr 2016, 12:28



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Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]
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03 Apr 2016, 17:39
Chiragjordan wrote: Why is that ABEH & BDFH => Angle DBA = Angle FHE i mean when i look at the figure they sure seem equal But I couldn't figure that out while answering... Dear Chiragjordan, I'm happy to respond. I was the creator of this question. Think about it this way. Suppose we extended segment BCD so that it went off beyond D as a ray. Then, imagine that we extend segment HGE off as a ray beyond E. Those two rays would intersect: call that intersection point Q. You may have to draw the figure out on paper to visualize this. Think about the angle formed by those two rays at Q. Because ABEH, the long BCD ray is a transversal cutting those two parallel lines, so the angle at Q would equal the angle at B: they would be what are known in Geometry as "Alternate Interior Angles." Because BDFH, the long HGE ray is a s a transversal cutting those two parallel lines, so the angle at Q would equal the angle at H. The angle at B and the angle at H both equal the angle at Q, so they have to equal each other. That's one way to see it. Another way is to think about it as follows. Suppose I am standing at B, and my friend is standing at H. When I look from B through point A to the horizon, and my friend looks from point H through point E to the horizon, we both will be facing in the same direction, facing the same point on the horizon. Now, we both turn clockwise. After the turn, I am still at B, looking out through point D to the horizon, and my friend is still at H, looking out through point F to the horizon. We are still facing in the same direction, facing the same point on the horizon. Well, if we started facing the same direction, and then each turned and wound up facing the same direction, we must have turned the same amount. Therefore, the angle of our turning, the angle at B and the angle at H, must be equal. Yet another way to think about it would be: suppose we dropped the whole diagram into the Coordinate plane. ABEH, so they have the same slope: call that m1. BDFH, so they also have the same slope: call that m2. Well clearly, if two lines at two different slopes meet at some angle, then changing the yintercepts of those lines would move the intersection point around, but as along as the slopes are the same, the angle between the two lines are the same. Thus, if we look at the intersection of any line with a slope of m1 with any line with a slope of m2, they will intersect at the same angle. That's exactly what we have with angle B and angle H. Does all this make sense? Mike
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Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]
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03 Apr 2016, 22:41
mikemcgarry wrote: Chiragjordan wrote: Why is that ABEH & BDFH => Angle DBA = Angle FHE i mean when i look at the figure they sure seem equal But I couldn't figure that out while answering... Dear Chiragjordan, I'm happy to respond. I was the creator of this question. Think about it this way. Suppose we extended segment BCD so that it went off beyond D as a ray. Then, imagine that we extend segment HGE off as a ray beyond E. Those two rays would intersect: call that intersection point Q. You may have to draw the figure out on paper to visualize this. Think about the angle formed by those two rays at Q. Because ABEH, the long BCD ray is a transversal cutting those two parallel lines, so the angle at Q would equal the angle at B: they would be what are known in Geometry as "Alternate Interior Angles." Because BDFH, the long HGE ray is a s a transversal cutting those two parallel lines, so the angle at Q would equal the angle at H. The angle at B and the angle at H both equal the angle at Q, so they have to equal each other. That's one way to see it. Another way is to think about it as follows. Suppose I am standing at B, and my friend is standing at H. When I look from B through point A to the horizon, and my friend looks from point H through point E to the horizon, we both will be facing in the same direction, facing the same point on the horizon. Now, we both turn clockwise. After the turn, I am still at B, looking out through point D to the horizon, and my friend is still at H, looking out through point F to the horizon. We are still facing in the same direction, facing the same point on the horizon. Well, if we started facing the same direction, and then each turned and wound up facing the same direction, we must have turned the same amount. Therefore, the angle of our turning, the angle at B and the angle at H, must be equal. Yet another way to think about it would be: suppose we dropped the whole diagram into the Coordinate plane. ABEH, so they have the same slope: call that m1. BDFH, so they also have the same slope: call that m2. Well clearly, if two lines at two different slopes meet at some angle, then changing the yintercepts of those lines would move the intersection point around, but as along as the slopes are the same, the angle between the two lines are the same. Thus, if we look at the intersection of any line with a slope of m1 with any line with a slope of m2, they will intersect at the same angle. That's exactly what we have with angle B and angle H. Does all this make sense? Mike Superb Explanation mikemcgarry ; the only thing that bothers me here is => It still taking me approximately 3 minutes to solve this question.. Wonder what will happen during the Actual Gmat.. Having said that => This is one heck of a question.. Wait ...lemme tap that Kudos.. Thanks
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Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]
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21 Sep 2017, 03:28
missionphd wrote: ABEH and AG is transversal => Angle EGA = Angle GAB = 70 degree [ Alternate Angles]
Angle GAB = Angle CAB = Angle ACB = 70 (AB = CB) => Angle B = 40
again, ABEH & BDFH => Angle DBA = Angle FHE = 40 degree
Therefore, Angle EGH = (180  2*40) = 100 degrees
Now, BDFH and DF is transversal, so Angle BDF + Angle HFE= 180 => BDF + 100 = 180 => CDE = 80 Answe : D can somebody please let me know .. how come. Angle DBA = Angle FHE = 40 degree




Re: In the diagram above, AB is parallel to EH, and BD is parall
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