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Geometry Problem

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Director
Joined: 10 Sep 2007
Posts: 938

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20 Jul 2008, 09:50
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Please see the attached document for the figure. R1 bisects angle ACB and R2 bisects angle CAD. Is R1 parallel to R2?

1) AB = DC
2) BD bisects AC.
Attachments

Parallel_Jumble.JPG [ 23.22 KiB | Viewed 1406 times ]

Director
Joined: 23 Sep 2007
Posts: 783

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20 Jul 2008, 10:02
statement 1: the 2 segments are equal, but not necessary parallel
insuf

statement 2: BD bisects AC ---> translate to parallel lines
suff

B
Director
Joined: 01 Jan 2008
Posts: 622

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20 Jul 2008, 10:15
E is the answer. Let's draw B close to A (AB is relatively small, say equal to 1) and C is far away from both A and B, for example AC = 10. Then we can draw a line through B and O, the middle of AC. If D belongs to that line than BD bisects AC. Let's draw a circle with the center in C and radius AB, it intersects line BO in two points. One of those points (D') gives parallel lines R1 and R2 because angle BCA = CAD', the other one doesn't because BCA is not equal to CAD''.
Director
Joined: 10 Sep 2007
Posts: 938

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20 Jul 2008, 10:29
As per OA, both of you had it wrong. I will post the OA after someone post the correct approach.
Senior Manager
Joined: 07 Jan 2008
Posts: 289

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20 Jul 2008, 10:34
I believe its a combination of 1) and 2) that makes the figure a prallelogram. Agreed?
If yes then yes these lines should be parallel.
C.
Director
Joined: 01 Jan 2008
Posts: 622

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20 Jul 2008, 10:37
mbawaters wrote:
I believe its a combination of 1) and 2) that makes the figure a prallelogram. Agreed?
If yes then yes these lines should be parallel.
C.

I disagree. There is no guarantee that it's going to be a parallelogram, solution is not unique.
Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12

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20 Jul 2008, 11:44
i agree we dont know of the relationship btw R1 and R2...

I too would guess E on this..
Manager
Joined: 04 Apr 2008
Posts: 215
Location: Pune

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20 Jul 2008, 20:39
even i will go with E
_________________

Every Problem Has a Sloution So keep working
AB

Senior Manager
Joined: 07 Jan 2008
Posts: 289

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21 Jul 2008, 05:18
maratikus wrote:
mbawaters wrote:
I believe its a combination of 1) and 2) that makes the figure a prallelogram. Agreed?
If yes then yes these lines should be parallel.
C.

I disagree. There is no guarantee that it's going to be a parallelogram, solution is not unique.

A good thing to know here is that what is a minimum requirement for a quadrilateral to be a parallelogram? I thought having diagnolas bisecting and two of the opposite sides being equal was good enough a reason to assume this quadrilateral was a parallelogram.
Manager
Joined: 15 Jul 2008
Posts: 206

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21 Jul 2008, 05:36
I also thought the answer is B.

Bisecting diagonals is sufficient condition for || gm. BD and AC bisect and so ABCD is ||gm.

now for 2 lines R1 and R2, if the alternate angels formed by the transveral(AC) are equal, then R1 and R2 are parallel.

Wonder why not B then.
Director
Joined: 01 Jan 2008
Posts: 622

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21 Jul 2008, 06:14
mbawaters wrote:
A good thing to know here is that what is a minimum requirement for a quadrilateral to be a parallelogram? I thought having diagnolas bisecting and two of the opposite sides being equal was good enough a reason to assume this quadrilateral was a parallelogram.

It says BD bisects AC. It doesn't say anywhere that BD and AC bisect each other.
Director
Joined: 10 Sep 2007
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21 Jul 2008, 06:58
OA is C. Even I went for E. It does not make any sense to me either.
Senior Manager
Joined: 07 Jan 2008
Posts: 289

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21 Jul 2008, 07:22
maratikus wrote:
mbawaters wrote:
A good thing to know here is that what is a minimum requirement for a quadrilateral to be a parallelogram? I thought having diagnolas bisecting and two of the opposite sides being equal was good enough a reason to assume this quadrilateral was a parallelogram.

It says BD bisects AC. It doesn't say anywhere that BD and AC bisect each other.

Sure! but we have two properties here

1) AB = DC
2) BD bisects AC

Can AB be equal to DC if AC also did not bisect BD? I was not able to come up with a case...
Manager
Joined: 15 Jul 2008
Posts: 206

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21 Jul 2008, 07:43
maratikus wrote:
It says BD bisects AC. It doesn't say anywhere that BD and AC bisect each other.

Good catch there.. such things really hurt...
Director
Joined: 01 Jan 2008
Posts: 622

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21 Jul 2008, 12:00
mbawaters wrote:
maratikus wrote:
mbawaters wrote:
A good thing to know here is that what is a minimum requirement for a quadrilateral to be a parallelogram? I thought having diagnolas bisecting and two of the opposite sides being equal was good enough a reason to assume this quadrilateral was a parallelogram.

It says BD bisects AC. It doesn't say anywhere that BD and AC bisect each other.

Sure! but we have two properties here

1) AB = DC
2) BD bisects AC

Can AB be equal to DC if AC also did not bisect BD? I was not able to come up with a case...

E is the answer. Let's draw B close to A (AB is relatively small, say equal to 1) and C is far away from both A and B, for example AC = 10. Then we can draw a line through B and O, the middle of AC. If D belongs to that line than BD bisects AC. Let's draw a circle with the center in C and radius AB, it intersects line BO in two points. One of those points (D') gives parallel lines R1 and R2 because angle BCA = CAD', the other one doesn't because BCA is not equal to CAD''.
Intern
Joined: 22 Jul 2008
Posts: 5

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22 Jul 2008, 20:24
mbawaters wrote:

1) AB = DC
2) BD bisects AC

Can AB be equal to DC if AC also did not bisect BD? I was not able to come up with a case...

Yes. AB can equal DC while AC does not bisect BD. Consider a trapezoid with the following dimensions AB=1, BC=1, CD=1, DA=2.
Intern
Joined: 22 Jul 2008
Posts: 5

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23 Jul 2008, 04:40
Statement 2, by itself, tells us that AB must equal BC and that AD must equal CD. Imagining a kite (in which the vertical axis bisects the horizontal axis (but not vice-versa)) may make the rule easier to understand. Combining this conclusion with statement 1 indicates that all four segments in the perimeter must be equal (transitive rule of equalities), therefore, the figure is a rhombus (but not necessarily a square). Rhombuses are a class of parallelogram, so AC then becomes a transversal of two parallel lines so angle BCA = angle CAD.

Re: Geometry Problem   [#permalink] 23 Jul 2008, 04:40
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