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In the figure, ABCD is a square and E is a point inside the square. Is

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In the figure, ABCD is a square and E is a point inside the square. Is [#permalink]

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New post 12 May 2017, 13:10
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In the figure, ABCD is a square and E is a point inside the square. Is BEC a right triangle?


(1) The point E is on the line joining the midpoints of AD and BC.

(2) AED is 100°.


Source: Nova GMAT

Attachment:
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Untitled.jpg [ 20.27 KiB | Viewed 1961 times ]

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Re: In the figure, ABCD is a square and E is a point inside the square. Is [#permalink]

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New post 13 May 2017, 04:36
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1
SajjadAhmad wrote:
In the figure, ABCD is a square and E is a point inside the square. Is BEC a right triangle?

(1) The point E is on the line joining the midpoints of AD and BC.
(2) AED is 100°.

Source: Nova GMAT



Hi,

Let's see the statements..

(1) The point E is on the line joining the midpoints of AD and BC.
This tells us that point E is equidistant from A and D and also from B and C.
There is ONLY one position where BEC will be 90° and that is at the point where all four A,B,C and D are Equidistant. This point will be also where the diagonals meet/bisect each other. At all other locations it will NOT be 90°

Insufficient

(2) AED is 100°.
There will be many places inside where this is possible. At some it can be 90°
Insufficient

Combined
We know the line on which this point exists and at only one place BEC will be 90°..
And this is the point T where diagonals intersect and all angles are 90° here..
But AED is not 90° so BEC will not be 90°

Ans is No always
Sufficient

C
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Re: In the figure, ABCD is a square and E is a point inside the square. Is [#permalink]

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New post 17 May 2017, 03:53
1
chetan2u wrote:
SajjadAhmad wrote:
In the figure, ABCD is a square and E is a point inside the square. Is BEC a right triangle?

(1) The point E is on the line joining the midpoints of AD and BC.
(2) AED is 100°.

Source: Nova GMAT



Hi,

Let's see the statements..

(1) The point E is on the line joining the midpoints of AD and BC.
This tells us that point E is equidistant from A and D and also from B and C.
There is ONLY one position where BEC will be 90° and that is at the point where all four A,B,C and D are Equidistant. This point will be also where the diagonals meet/bisect each other. At all other locations it will NOT be 90°

Insufficient

(2) AED is 100°.
There will be many places inside where this is possible. At some it can be 90°
Insufficient

Combined
We know the line on which this point exists and at only one place BEC will be 90°..
And this is the point T where diagonals intersect and all angles are 90° here..
But AED is not 90° so BEC will not be 90°

Ans is No always
Sufficient

C



Regarding the first statement, if we were told E was on the midpoint of the line joining the 2 midpoints (disregard the second statement for now), that would make point E the center and thus equidistant from all the points on the square, right?
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Re: In the figure, ABCD is a square and E is a point inside the square. Is [#permalink]

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New post 25 May 2017, 07:34
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From Statement (1) alone, we have that point E is on the line joining the midpoints of AD and BC. Since there are infinitely many points on the line joining the midpoints and each point makes a different BEC ranging between approximately 17° (when E is on AD) and 180° (when E is on BC), BEC need not be a right angle. Hence, BEC may or may not be a right angle. Hence, Statement (1) alone is not sufficient.
Now, from Statement (2) alone, we have that AED is 100°. Hence, E makes an angle of 100° with any point in the points A and D. Hence, E could be any point on the circular arc made by AD as a chord and with angle of the chord equal to 100° as shown in the figure. Each point makes a different angle BEC starting from 0° to 180°. Since the range includes the angle 90°, the triangle may or may not be a right triangle.
Now, with the statements together, the point E is the common point between the circular arc and the line joining the midpoints. There is only one such point and therefore, since we know the unique such point, we can always measure the angle made by the BEC. We are not interested in the actual measure. We are only interested in whether we have data sufficient to answer the question. Hence, the statements together answer the question, and therefore the answer is (C)
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Re: In the figure, ABCD is a square and E is a point inside the square. Is [#permalink]

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New post Updated on: 16 Jun 2018, 02:02
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Originally posted by gmatbusters on 08 Apr 2018, 22:34.
Last edited by gmatbusters on 16 Jun 2018, 02:02, edited 2 times in total.
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Re: In the figure, ABCD is a square and E is a point inside the square. Is [#permalink]

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New post 21 Apr 2018, 00:50
chetan2u wrote:
SajjadAhmad wrote:
In the figure, ABCD is a square and E is a point inside the square. Is BEC a right triangle?

(1) The point E is on the line joining the midpoints of AD and BC.
(2) AED is 100°.

Source: Nova GMAT



Hi,

Let's see the statements..

(1) The point E is on the line joining the midpoints of AD and BC.
This tells us that point E is equidistant from A and D and also from B and C.
There is ONLY one position where BEC will be 90° and that is at the point where all four A,B,C and D are Equidistant. This point will be also where the diagonals meet/bisect each other. At all other locations it will NOT be 90°

Insufficient

(2) AED is 100°.
There will be many places inside where this is possible. At some it can be 90°
Insufficient

Combined
We know the line on which this point exists and at only one place BEC will be 90°..
And this is the point T where diagonals intersect and all angles are 90° here..
But AED is not 90° so BEC will not be 90°

Ans is No always
Sufficient

C

Hi chetan2u

I am wondering whether B alone would not suffice i.e. there is no way for BEC to 90 degrees if AED is 100 degrees. I would like to prove myself wrong. Would you please see any way of drawing a configuration where AED is 100 degrees AND BEC is 90 degrees?

Thank you
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Re: In the figure, ABCD is a square and E is a point inside the square. Is [#permalink]

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New post 21 Apr 2018, 02:15
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1
You are right, I fell in the trap and marked C. :sad:

Explanation for statement 2: The Angle BEC shall be 90 deg only if E point lies on the Semi-Circle on BC as diameter.

Now Similarly, on same logic, AED will be 100 if it is inside the Semicircle on AD as diameter.

Please see the Sketch.
Attachment:
IMG_20180421_144023.jpg
IMG_20180421_144023.jpg [ 713.13 KiB | Viewed 543 times ]


Hence we can’t ever have angle AED = 100 and angle BEC = 90.

So statement 2 is sufficient.

Answer should be B.

Bunuel, chetan2u. Please approve my approach.


Octobre wrote:
chetan2u wrote:
SajjadAhmad wrote:
In the figure, ABCD is a square and E is a point inside the square. Is BEC a right triangle?

(1) The point E is on the line joining the midpoints of AD and BC.
(2) AED is 100°.

Source: Nova GMAT



Hi,

Let's see the statements..

(1) The point E is on the line joining the midpoints of AD and BC.
This tells us that point E is equidistant from A and D and also from B and C.
There is ONLY one position where BEC will be 90° and that is at the point where all four A,B,C and D are Equidistant. This point will be also where the diagonals meet/bisect each other. At all other locations it will NOT be 90°

Insufficient

(2) AED is 100°.
There will be many places inside where this is possible. At some it can be 90°
Insufficient

Combined
We know the line on which this point exists and at only one place BEC will be 90°..
And this is the point T where diagonals intersect and all angles are 90° here..
But AED is not 90° so BEC will not be 90°

Ans is No always
Sufficient

C

Hi chetan2u

I am wondering whether B alone would not suffice i.e. there is no way for BEC to 90 degrees if AED is 100 degrees. I would like to prove myself wrong. Would you please see any way of drawing a configuration where AED is 100 degrees AND BEC is 90 degrees?

Thank you

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Re: In the figure, ABCD is a square and E is a point inside the square. Is [#permalink]

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New post 21 Apr 2018, 20:53
A simpler approach: since ABCD is a square, hence the BED will be 90 deg only at the point of intersection of diagonals which makes angle AED also 90 deg. Since angle AED is 100 deg. It is sufficient to tell that BED is not a right angled triangle.

Hence B
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Re: In the figure, ABCD is a square and E is a point inside the square. Is [#permalink]

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New post 16 Jun 2018, 01:57
SajjadAhmad wrote:
In the figure, ABCD is a square and E is a point inside the square. Is BEC a right triangle?

(1) The point E is on the line joining the midpoints of AD and BC.
(2) AED is 100°.

Source: Nova GMAT



Hi Bunuel ....Just wanted to confirm if the correct answer is "B"..Going by the discussion thread it seems like B but official ans is marked as C.
Kindly confirm.. Thanks
Re: In the figure, ABCD is a square and E is a point inside the square. Is   [#permalink] 16 Jun 2018, 01:57
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