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Quadrilateral ABCD is a rhombus and points C, D, and E are [#permalink]
 09 Aug 2009, 10:03
Question Stats:
36% (02:35) correct
63% (01:14) wrong based on 12 sessions
Attachment: 
Rhombus.gif [ 1.37 KiB | Viewed 9860 times ]
Quadrilateral ABCD is a rhombus and points C, D, and E are on the same line. Is quadrilateral ABDE a rhombus? (1) The measure of angle BCD is 60 degrees. (2) AE is parallel to BD
Last edited by Bunuel on 19 Jan 2013, 04:44, edited 1 time in total.
OA added.
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Re: Rhombus hard problem [#permalink]
09 Aug 2009, 10:14
[quote="crejoc"]Quadrilateral ABCD is a rhombus and points C, D, and E are on the same line. Is quadrilateral ABDE a rhombus?
(1) The measure of angle BCD is 60 degrees.
(2) AE is parallel to BD
to prove that it is a rhombus, we need to prove that it is a paralellogram with equal opposite angles and all sides =.
from 1
draw the diagonal bd would split the abcd rhombus into 2 similar triangles , both eqelateral all angles = 60, however as long as we dont know whether ae is // to bd or we know angles dae or aed we can not deduce that opposit sides of abde are equal or parallel.....insuff
from 2
obviously not suff
both
suff...C
Last edited by yezz on 15 Aug 2009, 12:05, edited 1 time in total.
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Re: Rhombus hard problem [#permalink]
15 Aug 2009, 12:08
sandipchowdhury wrote: whould you please explain why not B ? to prove a shape to be a rhombus:1)opposite sides are // 2) all sides are equal and ( only to deferenciate it from a square): 3) opposite angles are =
to prove to be a squaresame as above however all angles have to be = in measure and a such each = 90 degrees
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Re: Rhombus hard problem [#permalink]
22 Oct 2009, 06:48
I don't understand why not B. question stem has given CDE is parallel to AB --> DE is parallel to AB and S2 give AE is parallel to DB , so for ABDE, we have the condition that opposite sides are parallel is met. How we know that opposite angles are not equal - can someone draw such figure ?
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Re: Rhombus hard problem [#permalink]
25 Oct 2009, 04:42
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'B' meets only one condition for a rhombus which is Parellelism. but it does not prove that all sides of ABDE are equal. Please see the attachment which suggests that 'B' only is not correct. We also need 'A' to prove all sides are equal. Hope it clarifies.
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Re: Rhombus hard problem [#permalink]
25 Oct 2009, 13:56
puneet478 wrote: 'B' meets only one condition for a rhombus which is Parellelism. but it does not prove that all sides of ABDE are equal. Please see the attachment which suggests that 'B' only is not correct. We also need 'A' to prove all sides are equal. Hope it clarifies. Thanks puneet, that helps...
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Quadrilateral ABCD is a rhombus and points C, D, and E are on the same line. Is quadrilateral ABDE a rhombus?Rhombus is a quadrilateral with all four sides equal in length. A rhombus is actually just a special type of parallelogram (just like square or rectangle). So ABCD is a rhombus means AB=BC=CD=AD. ABDE to be a rhombus it must be true that AB=BD=DE=AE. (1) The measure of angle BCD is 60 degrees --> diagonal BD equals to the sides of rhombus, so BD=AB. Know nothing about DE or/and AE. Not sufficient. (2) AE is parallel to BD --> ABDE is a parallelogram (as AE||BD and BA||DE), hence opposite sides are equal: BD=AE and AB=DE. But we don't know whether all sides are equal (AB=BD=DE=AE). Not sufficient. (1)+(2) From (1): BD=AB and from (2) BD=AE and AB=DE --> AB=BD=DE=AE --> ABDE is a rhombus. Sufficient. Answer: C.
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AE is parallel to BD --> ABDE is a parallelogram (as AE||BD and BA||DE), hence opposite sides are equal: BD=AE and AB=DE. But we don't know whether all sides are equal (AB=BD=DE=AE)
Bunuel, all sides have to be equal as the question stem states that C, D and E are on the same line. And it also states that BD is parallel to AE. Try drawing any kind of rhombus with the following conditions and all sides will be equal. So why do we need statement A? Am I missing something?
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study wrote: AE is parallel to BD --> ABDE is a parallelogram (as AE||BD and BA||DE), hence opposite sides are equal: BD=AE and AB=DE. But we don't know whether all sides are equal (AB=BD=DE=AE)
Bunuel, all sides have to be equal as the question stem states that C, D and E are on the same line. And it also states that BD is parallel to AE. Try drawing any kind of rhombus with the following conditions and all sides will be equal. So why do we need statement A? Am I missing something? From your reasoning above it's not clear how you came to the conclusion that alls sides must be equal. Actually I don't even need to try drawing, to state that there are infinite # of cases possible for AE to be parallel to BD and ABDE not to be a rhombus. Just try to increase or decrease diagonal BD and leave everything else the same (AE||BD): you'll always have a parallelogram but in only one case a rhombus, when BD=AB.
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Bunuel, Thanks!! makes sense.
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Re: Rhombus hard problem [#permalink]
16 Oct 2010, 09:35
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anandnat wrote: I still don't understand why B is wrong. Can we safely say that the diagonal will never equal the side? If this is true, then with B, we always get a firm answer that ABDE is never a rhombus. Hence imo the answer is B. Math experts please help! anandnat, With statement 2, we can conclude that since AE is parallel to BD, therefore triangle ABD is mirror image of AED (similar triangle). We have, AD is equal to AB. With all this, we can assert that ED is equal to AB and AE is equal to BD. In other way, to cut the long story short:- From st 2, we can come closer to only this much..  .. ABD and AED are two similar "Isosceles" triangles, Joined together. But, we need to prove that all four sides are equal. I have drawn one such example here: Hope that helps!
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Last edited by samark on 17 Oct 2010, 04:29, edited 1 time in total.
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Re: Rhombus hard problem [#permalink]
17 Oct 2010, 03:57
Thanks samark, great explanation. +1
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Re: Rhombus hard problem [#permalink]
19 Oct 2010, 20:25
I will explain why B) is not sufficient Pls refer the attached diagram AE || BD is not sufficient to judge whether BD = DC or BD = BC because for AEDB to be a rhombus , AE = ED = DB= BA From A) we can deduce DC = CB = BD (diagonal of the rhombus) as angle B = angle C = angle D = 60 deg. Now from the statement of the question , DC = AB as it is a rhombus , so DB = AB Since from A) we deduce DB = AB and from 2) we know that AE = DB as C , D , E lies on a straight line Hence combining (A) and (B) , we know that AEDB is always a RHOMBUS Note that if angle BCD not equal to 60 degrees , then AEDB would not have been a RHOMBUS Hope the above explanation is now clear
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Re: Rhombus hard problem [#permalink]
20 Oct 2010, 05:48
good explanation Bunuel..txs
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Re: Quadrilateral ABCD is a rhombus and points C, D, and E are [#permalink]
25 Mar 2012, 17:58
Bunuel, thank you. great explanation.
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Re: Quadrilateral ABCD is a rhombus and points C, D, and E are [#permalink]
10 May 2012, 14:47
I still don't understand this question. "AE is parallel to BD --> ABDE is a parallelogram (as AE||BD and BA||DE), hence opposite sides are equal: BD=AE and AB=DE. " I understand how paralellogram --> BD=AE and AB=DE, but how does AE||BD and BA||DA imply it is a paralellogram with opposite sides equal?
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Re: Quadrilateral ABCD is a rhombus and points C, D, and E are [#permalink]
18 Jan 2013, 20:51
Can some1 pls explain me how from st1 people derive BD=AB.Can anyone pls explain elaborately
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Re: Quadrilateral ABCD is a rhombus and points C, D, and E are [#permalink]
19 Jan 2013, 04:54
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Bunuel wrote: Attachment: untitled.JPG Quadrilateral ABCD is a rhombus and points C, D, and E are on the same line. Is quadrilateral ABDE a rhombus?Rhombus is a quadrilateral with all four sides equal in length. A rhombus is actually just a special type of parallelogram (just like square or rectangle). So ABCD is a rhombus means AB=BC=CD=AD. ABDE to be a rhombus it must be true that AB=BD=DE=AE. (1) The measure of angle BCD is 60 degrees --> diagonal BD equals to the sides of rhombus, so BD=AB. Know nothing about DE or/and AE. Not sufficient. (2) AE is parallel to BD --> ABDE is a parallelogram (as AE||BD and BA||DE), hence opposite sides are equal: BD=AE and AB=DE. But we don't know whether all sides are equal (AB=BD=DE=AE). Not sufficient. (1)+(2) From (1): BD=AB and from (2) BD=AE and AB=DE --> AB=BD=DE=AE --> ABDE is a rhombus. Sufficient. Answer: C. Bunnel, In Statement 2 How can you say ABDE is ||gm without knowing whether AB and DE are ||el.... we just know that AE and BD are ||el
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