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mads
If ABCD is a quadrilateral,
Statement 1: Line segments AC and BD bisect one another. This is the case for rectangle, square and rhombus.
Not suff.
Stetement 2: Angle ABC is a right angle.
Any quadrilateral can have one right angle and other different angle. Not Suff.

Combining both statements could give us either rectangle or a square. Not suff.

But isn't square a specific case of rectangle. That way i would say answer is C
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OK! What if it is a right trapezoid? I mean it will have one (even two) right angles, the segments AC and BD will bisect one another and at the same time it will not be a rectagle. What in this case?
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@ mossovet814: If it is a trapezoid then the diagonals (AC and BD) won't bisect each other.

Answer should be C.
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pankajattri
@ mossovet814: If it is a trapezoid then the diagonals (AC and BD) won't bisect each other.

Answer should be C.

OK! This is a right trapezoid. Two angles are right and diagonals bisect one another. Still it is not a rectangle. I know the question is stupid... but I just want to find a flaw in my reasoning.
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Financier
Ok I must admit I am confused. Diagonals of a trapezoid bisect each other?
a). Diagonals of quadrangle allways bisect each other.
b). Trapezoid is a specific case of quadrangle .
I don't think that the diagonals of a trapezoid bisect each other.

Financier
pankajattri
@ mossovet814: If it is a trapezoid then the diagonals (AC and BD) won't bisect each other.

Answer should be C.


Dear Pankajattri,


First of all, I strongly suggest that you review the basics of Geometry so that you could recall that:
a). Diagonals of quadrangle allways bisect each other.
b). Trapezoid is a specific case of quadrangle .

Second, the Rules of GMATclub suppose that folks do not just utter statements without any foundation,
but rather unveil what rules they are based on. Above are the statements that prove that diagonals of Trapezoid intersect each other.

Third, the correct answer is E.
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Dear Financier,

Even If I was wrong, which by the way I am not, I don't know why couldn't you find a more suitable way of posting your comment.

As for as the question is concerned; the diagonals of a PARALLELOGRAM bisect each other and not of each QUADRANGLE. All parallelograms are quadrangles but not vice versa.

I really hope that you know what does "bisection", "quadrangle" and "parallelogram" mean.

And next time please do not utter statements without any foundation.
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Pankajattri,

I was wrong, excuse me and my tone. I'm soooo sorry. I messed up words "bisect" and "intersect". This happens when people study a lot. The correct answer is "C".
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Guys just wanted to add . as I had little confusion when I first saw this question --

From (1) ----> The diagonals bisect each other , ----> its a ||gm , for it to be a recrangle both diagonals should be equal.
From(2) ----> |_ ABC= 90 ----> clearly not sufficient.

When we combine both 1 & 2 ----> Its a ||gm with one angle 90. ----> that has to be a rectangle and both diagonals become automatically equivalent.
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I don't understand this.

Question: IS ABCD rectangle?

If we take "C", couldn't it still be a square?? Or is a square a "special rectangle with 4 equal sides"?

Thanks for you help.

Cheers
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agautam7879
Is quadrilateral ABCD a rectangle?
(1) Line segments AC and BD bisect one another.

(2) Angle ABC is a right angle.

For me the answer should be E...because the ABCD could be a square as well...please comment

option B does not tell us whether all angles are 90. How option C is correct :(
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agautam7879
Is quadrilateral ABCD a rectangle?
(1) Line segments AC and BD bisect one another.

(2) Angle ABC is a right angle.

For me the answer should be E...because the ABCD could be a square as well...please comment

option B does not tell us whether all angles are 90. How option C is correct :(

From (1) ABCD is a rectangle or a parallelogram. A parallelogram with a right angle is a rectangle.
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Can anyone articulate how we deduce from the fact that, if one angle of a parallelogram is 90 degrees and the opposite angle is equal, how do we come to the conclusion that the other pair of opposite angles are also equal to 90 degrees?
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Can anyone articulate how we deduce from the fact that, if one angle of a parallelogram is 90 degrees and the opposite angle is equal, how do we come to the conclusion that the other pair of opposite angles are also equal to 90 degrees?

Useful property: the adjacent angles of a parallelogram are supplementary (supplementary angles are two angles that add up to 180°). So, if angle B is 90°, then A and C must also be 90° to add up to 180.

Hope it helps.
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Bunuel @chentan2u cant it be Rhombus? Because rhombus diagonals bisect each other
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Bishal123456789
Bunuel @chentan2u cant it be Rhombus? Because rhombus diagonals bisect each other

Yes, but a rhombus is a parallelogram (A rhombus is a parallelogram with four congruent sides).
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Statement 1 tells us that the shape is a parallelogram, but it could be a square, rhombus, or rectangle

Statement 2 tells us that the shape has 1 90 degree angle.

Statement 1 + 2 tells us that it is a parallelogram that has 1 90 degree angle, so we can deduce the following:
- Other angles must also be 90 degrees as angles in parallelograms are supplementary
- the shape cannot be a rhombus as rhombi do not have 90 degree angles
This allows us to conclude that the shape is either a square or rectangle.
In GMATLand it is common to get tested on the trap that a square is actually a form of rectangle so keep this in mind.

Combined = sufficient.
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Bishal123456789
Bunuel @chentan2u cant it be Rhombus? Because rhombus diagonals bisect each other

Yes, but a rhombus is a parallelogram (A rhombus is a parallelogram with four congruent sides).
Bunuel

A rhombus is a parallelogram but in rhombus diagnols bisect at 90 but not in rectangle.
So how can we say it is rectangle Nd not rhombus.
Every square is rectangle and every square is rhombus , but not vice vers.
Is same relation true for rectangle .?
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