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In the figure above, D is a point on the side AC of ΔABC.Is ΔABC isos?

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In the figure above, D is a point on side AC of ΔABC. Is ΔABC is isosc  [#permalink]

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New post 09 Dec 2014, 15:39
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Tough and Tricky questions: Geometry.



Image
In the figure above, D is a point on side AC of ΔABC. Is ΔABC is isosceles?

(1) The area of triangular region ABD is equal to the area of triangular region DBC.
(2) BD┴AC and AD = DC


Kudos for a correct solution.

Attachment:
triang ABCD.JPG
triang ABCD.JPG [ 3.38 KiB | Viewed 3821 times ]

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Re: In the figure above, D is a point on side AC of ΔABC. Is ΔABC is isosc  [#permalink]

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New post 09 Dec 2014, 20:53
1
Statement 1: The area of triangular region ABD is equal to the area of triangular region DBC.

This means that the point D is mid-point of AC. But, we can't still say whether the triangle ABC is isosceles or not. Insufficient.

Statement 2: BD┴AC and AD = DC. Signifies that the median drawn from the vertex B is a perpendicular bisector. This can only happen if the triangle is isosceles triangle with AB = BC. Sufficient

The answer should be B
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Re: In the figure above, D is a point on side AC of ΔABC. Is ΔABC is isosc  [#permalink]

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New post 10 Dec 2014, 05:24
In the figure above, D is a point on side AC of ΔABC. Is ΔABC is isosceles?

(1) The area of triangular region ABD is equal to the area of triangular region DBC --> this gives us info that D is the midpoint of AC but we don't have any info about angle ADB & BDC....insufficient
(2) BD┴AC and AD = DC --> this gives us info that D is midpoint of AC and that BD┴AC, this condition will make AB=BC.......sufficient.(refer below image, hope my understanding is correct :? )

Ans. B)
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Re: In the figure above, D is a point on side AC of ΔABC. Is ΔABC is isosc  [#permalink]

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New post 26 Dec 2014, 11:01
Bunuel wrote:

Tough and Tricky questions: Geometry.



Image
In the figure above, D is a point on side AC of ΔABC. Is ΔABC is isosceles?

(1) The area of triangular region ABD is equal to the area of triangular region DBC.
(2) BD┴AC and AD = DC


Kudos for a correct solution.

Attachment:
triang ABCD.JPG


The correct answer is B.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: In the figure above, D is a point on side AC of ΔABC. Is ΔABC is isosc  [#permalink]

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New post 25 Jul 2015, 02:43
What exactly does this "BD┴AC" say?

I don't know what the symbol means.
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Re: In the figure above, D is a point on side AC of ΔABC. Is ΔABC is isosc  [#permalink]

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New post 25 Jul 2015, 02:44
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Re: In the figure above, D is a point on side AC of ΔABC. Is ΔABC is isosc  [#permalink]

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New post 25 Jul 2015, 05:00
4
Bunuel wrote:

Tough and Tricky questions: Geometry.



Image
In the figure above, D is a point on side AC of ΔABC. Is ΔABC is isosceles?

(1) The area of triangular region ABD is equal to the area of triangular region DBC.
(2) BD┴AC and AD = DC


Kudos for a correct solution.

Attachment:
The attachment triang ABCD.JPG is no longer available


Per statement 1, Area Triangle (BAD} = Area Triangle {BDC}

---> 0.5*BE*AD = 0.5*BE*CD ---> AD = CD . But no information about AB and BC. Thus this triangle may or may not be isosceles. Thus this statement is not sufficient.

Per statement 2, BD┴AC and AD = DC ---> Triangles BDA and BDC are congruent. ---> AB = BC ---> Triangle ABC is isosceles. Thus this statement is sufficient.

B is the correct answer.
Attachments

2015-07-25_7-55-54.jpg
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In the figure above, D is a point on the side AC of ΔABC.Is ΔABC isos?  [#permalink]

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New post 20 Feb 2016, 09:42
Attachment:
1.png
1.png [ 16.33 KiB | Viewed 2356 times ]


In the figure above, D is a point on the side AC of ΔABC. Is ΔABC isosceles?

(1) The area of the triangular region ABD is equal to the area of triangular region DBC.

(2) BD is perpendicular to AC and AD = DC.
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In the figure above, D is a point on the side AC of ΔABC.Is ΔABC isos?  [#permalink]

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New post 20 Feb 2016, 10:03
Nevernevergiveup wrote:
Attachment:
1.png


In the figure above, D is a point on the side AC of ΔABC. Is ΔABC isosceles?

(1) The area of the triangular region ABD is equal to the area of triangular region DBC.

(2) BD is perpendicular to AC and AD = DC.


Statement 1 is straightforward to see that it is NOT sufficient. The most this statement will do is to provide a scenario when we have AD=CD. But this does not yet answer the question asked.

Statement 2, is tricky as in from the figure provided, BD can either be to the left of A or between A and C or to the right of C. But in only 1 case (of between A and C) will you get AD=CD such that BD \(\perp\) AC. With this scenario, triangle ABC is indeed isosceles (by making triangles ABD and BCD congruent to each other by the SAS property ---> AB=BC) . Hence sufficient.

B is the correct answer.

Hope this helps.
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Re: In the figure above, D is a point on the side AC of ΔABC.Is ΔABC isos?  [#permalink]

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New post 20 Feb 2016, 18:46
Answer: B

We are after: Is ΔABC isosceles?

(1) Not sufficient. Using the formula for areas of triangle, we get


Area of ΔABD = Area of ΔABD
1/2 * AD * h = 1/2 * DC * h

h is the height for both triangles, if you drop a perpendicular from point B on AC. This will fall on the extended AC (if you can imagine it)

The above equation yields to:
AD = AC

But we want to know if ΔABC is isosceles. This information doesn't help so it is INSUFFICIENT.

(2) Sufficient

Since BD ⊥AC, and D is on AC, imagine D dividing AC into two sides i.e. AD =AC.

Now imagine two right angled Δs inside ABC being split in the middle by BD. These two right angled Δs have the same base and height so the hypotenuse will be same i.e. AB = AC
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Re: In the figure above, D is a point on the side AC of ΔABC.Is ΔABC isos?  [#permalink]

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