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voodoochild
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WholeLottaLove
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a+b

imagine a kite. In that the diagonals AC= BD means all the angles are equal too.
However, angle ABC<BCD means even if all sides are equal, diagonal BD> AC.

drawing this will help to understand it surely.

C
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I first thought both diagonals of a rhombus are equal but apparently the diagonal opposite to the wider angle is more in length than the other diagonal...

there for C

based on statement 2 we can conclude the two possibilities as Square and rhombus

so not sufficient

stament 1 is obviously not sufficient. Combining both the information we can conclude the figure as rhombus and with the angle gives we can determine which diagonal is more in length
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nice question!! for a min i forgot rhombus... :evil:
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voodoochild
If A, B, C and D form a quadrilateral. Is AC > BD ?

(1) Angle ABC < Angle BCD
(2) AB = BC= CD = DA

Is there a nice OE for this one?
From my understanding the question is asking whether the diagonals are equal right?
But, I'm afraid I didn't get the right answer. Will an expert please elaborate on this one?

Happy to provide some Kudos if needed
Cheers!
J :)
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WholeLottaLove
If A, B, C and D form a quadrilateral. Is AC > BD ?

(1) Angle ABC < Angle BCD

This tells us that ABCD is not a square or a rectangle because naturally, both have four angles each equaling 90 degrees. With a triangle, the leg across from the largest angle is the longest but does it work like that with a four sided figure? Apparently not because 1 is not sufficient.

(2) AB = BC= CD = DA

This tells us that ABCD is either a square or a rhombus. If the figure is a square, then both diagonals AC = BD. If the figure is a rhombus, then one diagonal will be longer than the other. Insufficient.

1+2) This tells us that all four sides are equal and that one angle (or in this case, pairs of angles) is greater than another. The only possible shape that has four equal sides and different interior angles is a Rhombus. AC > BD. Sufficient.

WholeLotta Love for you buddy! +1 Kudos

Cheers!
J :)
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voodoochild
If A, B, C and D form a quadrilateral. Is AC > BD ?

(1) Angle ABC < Angle BCD
(2) AB = BC= CD = DA

To show more clearly why 1 is not sufficient , here are 2 figures.
Both have angle B less than angle C

Hence 1 is insufficient.

Attachment:
Kite.jpg
Kite.jpg [ 119.23 KiB | Viewed 9650 times ]
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If points A , B , C , and D form a quadrilateral, is AC longer than BD ?

1. ∠ABC>∠BCD
2. AB=BC=CD=DA

Can you please help me out why the triangle property of the sides opposite to the greater angle has the greater length not applicable in statement A?
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ranajoy42
If points A , B , C , and D form a quadrilateral, is AC longer than BD ?

1. ∠ABC>∠BCD
2. AB=BC=CD=DA

Can you please help me out why the triangle property of the sides opposite to the greater angle has the greater length not applicable in statement A?


Hi,
I am merging your topic as the same has been discussed..
you can go through the solution..

Specific to your Q..
the angles are of two different triangles ABC and BCD, but the property of the sides opposite to the greater angle has the greater length is applicable in the same triangle..

Hope it helps
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DS:
If points A, B, C and D form a quardrilateral, is AC longer than BC?

Stmt 1: Angle ABC > Angle BCD.
Stmt 2: AB=BC=CD=DA.

Guys this question is already answered here: https://gmatclub.com/forum/m14-q32-quad ... 76743.html

But, I have a different question. When I started solving and drew a quadrilateral, I got confused what if the quadrilateral is drawn like ACBD? Coz this is a DS question and one needs to consider all the possibilities and in ACBD you can't have AC and BC as diagonals but they will be sides. Am I just thinking too far, and if yes, how should I limit myself from thinking this far? Thanks in advance.
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In quadrilateral \(ABCD\), is \(AC\) longer than \(BD\)?

Note that the question essentially asks whether, in quadrilateral \(ABCD\), diagonal \(AC\) is longer than diagonal \(BD\).

(1) \(\angle ABC \gt \angle BCD\).

Consider two kite-shaped figures, which give two different answers:



(2) \(AB = BC = CD = DA\).

The above statement implies that the quadrilateral is a rhombus. However, we can have \(AC > BD\), \(AC < BD\), or, in the case of a square, even \(AC = BD\). Not sufficient.

(1)+(2) Since from (2) the quadrilateral is a rhombus, then the fact that \(\angle ABC \gt \angle BCD\) implies that the side opposite \(\angle ABC\), which is \(AC\), is longer than the side opposite \(\angle BCD\), which is \(BD\). Sufficient.


Answer: C
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