Quadrilateral ABCD is a rhombus and points C, D, and E are

[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

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

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 ?

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.

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!

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

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?

Can some1 pls explain me how from st1 people derive BD=AB.Can anyone pls explain elaborately

(1) The measure of angle BCD is 60 degrees. Since given that BC=DC, then <DBC=<BDC --> <DBC+<BDC+<BCD=180 degrees --> x+x+60=180 --> x=60 degrees. We have that triangle BCD is equilateral, thus BD=BC=DC. We know that AB=BC=CD=AD, thus BD=AB.

Hope it's clear.

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

We know that points C, D, and E are on the same line and since CD||AB, then the same line DE is also parallel to AB.

Hope it's clear.

I think the rubber band technique is effective. If you can stretch a side or dimension and come up with different results, then the information is INSUFFICIENT.

1. If BCD is 60 then BAD is also 60. Then we are left with two angles from left to right with 120 each. Imagine a straight line cutting the rhombus in half horizontally, what we got are two equilateral triangles ABD and BCD. For ABDE to become a rhombus, AE, BD,DE, and AE must have equal sides. Imagine pulling the line CDE a little longer through pt. E, then we could distort the figure and come up with a non-rhombus quadrialeteral. We could push it back and we could estimate a rhombus.

INSUFFICIENT.

2. Now imagine your rhombus ABCD and make it narrower, this will make BD and AE's lengths shorter than the size of a side of rhombus ABCD. Imagine your rhombus a little wider and this will make BD and AE's lengths longer. By rubber band technique, we know that we are not sure if ABDE is a rhombus.

INSUFFICIENT.

Together:
We know that ABD and BCD are equilateral triangles forming rhombus ABCD. Thus, line BD would be equal to all the sides of the rhombus.
Now we know that BD and AE are parallel each other fixed by the bordering lines of BA and CDE. Hence, BD = AE.
All the sides of the rhombus are equal to BD then also to AE.

To close the deal, AB and DE must be equal to become a rhombus. Since AE and BD are two parallel lines with equal length then, we are certain that AB and DE are also equal in length.

Answer: C

Hi Bunuel,

I think the answer to this Q shd be B. My soln is as follows:

let angle BAD be X, angle ABC be Y. Therefore since ABCD is a rhombus, angle BCD will be X and angle CDA will be Y. Also all the sides are equal of this rhombus, i.e., AB=AD=CD=BC. Now draw BD. Further, in the Q it is given that CDE is a straight line, that means AB is parallel to CDE. Therefore, we can say that angle ADE is X (alternate angles). Now acc. to second stmt, AE is parallel to BD. Let angle DAE =Z. Consequently, angle ADB =Z (alternate angles). Then angle BDC = Y-Z and angle ABD= Z (because AB=AD). That means X+Z+Z = 180. Therefore, in triangle DAE, angle A = Z and angle D =X. From this, we can calculate that angle AED = Z. This means AD=DE. And therefore because triangle ABD is similar to traingle ADE, BD will also be equal to AE. Thus all sides are equal. And we do not need any specific angle value.

Please help! as to why B cant be the answer. GMAT in two days!!

First of all when making such posts please attach a diagram. It's hard to follow all that angles in your explanation.

As for your solution: where did you prove that BD is equal to AB? In rhombus all sides must be equal.

Thanx Bunuel, noticed my error... and thanx a ton fr quick reply too

How do you know that BD=AB from 1?? Is there some hidden calculations done in there? All I see is BCD=60, you don't know the measures of any other angles so it doesn't really tell you anything