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

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Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]
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AB||EH 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, AB||EH & BD||FH => Angle DBA = Angle FHE = 40 degree

Therefore, Angle EGH = (180 - 2*40) = 100 degrees

Now, BD||FH 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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what a nightmare the question is..tried multiple times to draw on my paper..
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Line diagram%2C two pairs of parallel lines.JPG [ 38.22 KiB | Viewed 12557 times ]

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Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]
Why is that AB||EH & BD||FH => 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...
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Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]
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Chiragjordan wrote:
Why is that AB||EH & BD||FH => 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 B-C-D so that it went off beyond D as a ray. Then, imagine that we extend segment H-G-E 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 AB||EH, 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 BD||FH, 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. AB||EH, so they have the same slope: call that m1. BD||FH, 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 y-intercepts 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]
mikemcgarry wrote:
Chiragjordan wrote:
Why is that AB||EH & BD||FH => 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 B-C-D so that it went off beyond D as a ray. Then, imagine that we extend segment H-G-E 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 AB||EH, 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 BD||FH, 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. AB||EH, so they have the same slope: call that m1. BD||FH, 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 y-intercepts 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]
missionphd wrote:
AB||EH 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, AB||EH & BD||FH => Angle DBA = Angle FHE = 40 degree

Therefore, Angle EGH = (180 - 2*40) = 100 degrees

Now, BD||FH 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
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In the diagram above, AB is parallel to EH, and BD is parall [#permalink]
The hardest part to recognize (and best part of this figure to “takeaway” from this problem):

Because DB is parallel to FH

And then the rays that shoot out from each line are also parallel ——-> AB is parallel to EH

The angle created by the rays and the original parallel lines will be Symmetrical and this EQUAL

Hence:

Angle < ABD = Angle < EHF

Knowing that, the rest comes down to sporting the fact that:

Angle < CAB and Angle <EGC are Equal Alternate Interior Angles = 70 degrees

From there it comes to “angle chasing” by using the Isosceles Triangles and getting the 3 measures of the quadrilateral in the middle of the figure

Angle <D = (360) - (70) - (70) - (140) =

80

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Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]
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Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]
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