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# Isosceles Triangle

Author Message
Manager
Joined: 29 Jul 2007
Posts: 182

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27 Nov 2007, 15:43
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

See diagram.

Thanks.
Attachments

isosceles triangle.JPG [ 12.27 KiB | Viewed 692 times ]

VP
Joined: 10 Jun 2007
Posts: 1439

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27 Nov 2007, 18:41
Skewed wrote:
See diagram.

Thanks.

B for me.

(1) is not sufficient because if both triangles have the same area, using the area formula, and given that both have the same height, D is the mid point of AC. However, knowing this you cannot tell if the triangle is isso.
INSUFFICIENT

(2) SUFFICIENT because using Pyt theorem, the hyp of each triangle must be the same.
Manager
Joined: 29 Jul 2007
Posts: 182

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27 Nov 2007, 20:12
bkk145 wrote:
Skewed wrote:
See diagram.

Thanks.

B for me.

(1) is not sufficient because if both triangles have the same area, using the area formula, and given that both have the same height, D is the mid point of AC. However, knowing this you cannot tell if the triangle is isso.
INSUFFICIENT

(2) SUFFICIENT because using Pyt theorem, the hyp of each triangle must be the same.

For I, if the areas are the same, then D is the midpoint of AC right. So if AD= CD, then BD would have bisected AC. So isn't BD an angle bisector of angle ABC as well?
Director
Joined: 09 Aug 2006
Posts: 755

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27 Nov 2007, 22:27
Skewed wrote:
bkk145 wrote:
Skewed wrote:
See diagram.

Thanks.

B for me.

(1) is not sufficient because if both triangles have the same area, using the area formula, and given that both have the same height, D is the mid point of AC. However, knowing this you cannot tell if the triangle is isso.
INSUFFICIENT

(2) SUFFICIENT because using Pyt theorem, the hyp of each triangle must be the same.

For I, if the areas are the same, then D is the midpoint of AC right. So if AD= CD, then BD would have bisected AC. So isn't BD an angle bisector of angle ABC as well?

Yes, according to stat 1, angle ABD = angle DBC. But that does not make the triangle isoceles. That's only one angle of triangle ABD = one angle of triangle DBC.

Stat 1:
Since areas are equal: 1/2 *AD * h = 1/2 DC*h or AD=DC. But we cannot prove similarity or another angle of both triangles equal using this. Insuff.

Stat 2:
angle ABD = angle DBC
angle BDA = angle BDC
Suff.

B
Manager
Joined: 29 Jul 2007
Posts: 182

### Show Tags

27 Nov 2007, 23:29
GK_Gmat wrote:
Skewed wrote:
bkk145 wrote:
Skewed wrote:
See diagram.

Thanks.

B for me.

(1) is not sufficient because if both triangles have the same area, using the area formula, and given that both have the same height, D is the mid point of AC. However, knowing this you cannot tell if the triangle is isso.
INSUFFICIENT

(2) SUFFICIENT because using Pyt theorem, the hyp of each triangle must be the same.

For I, if the areas are the same, then D is the midpoint of AC right. So if AD= CD, then BD would have bisected AC. So isn't BD an angle bisector of angle ABC as well?

Yes, according to stat 1, angle ABD = angle DBC. But that does not make the triangle isoceles. That's only one angle of triangle ABD = one angle of triangle DBC.

Stat 1:
Since areas are equal: 1/2 *AD * h = 1/2 DC*h or AD=DC. But we cannot prove similarity or another angle of both triangles equal using this. Insuff.

Stat 2:
angle ABD = angle DBC
angle BDA = angle BDC
Suff.

B

But doesn't the height or the altitude have to be perpendicular to the base? Thereby creating a right triangle - ADB = 90=CDB??
Manager
Joined: 01 Oct 2007
Posts: 87

### Show Tags

28 Nov 2007, 00:46
Skewed wrote:
GK_Gmat wrote:
Skewed wrote:
bkk145 wrote:
Skewed wrote:
See diagram.

Thanks.

B for me.

(1) is not sufficient because if both triangles have the same area, using the area formula, and given that both have the same height, D is the mid point of AC. However, knowing this you cannot tell if the triangle is isso.
INSUFFICIENT

(2) SUFFICIENT because using Pyt theorem, the hyp of each triangle must be the same.

For I, if the areas are the same, then D is the midpoint of AC right. So if AD= CD, then BD would have bisected AC. So isn't BD an angle bisector of angle ABC as well?

Yes, according to stat 1, angle ABD = angle DBC. But that does not make the triangle isoceles. That's only one angle of triangle ABD = one angle of triangle DBC.

Stat 1:
Since areas are equal: 1/2 *AD * h = 1/2 DC*h or AD=DC. But we cannot prove similarity or another angle of both triangles equal using this. Insuff.

Stat 2:
angle ABD = angle DBC
angle BDA = angle BDC
Suff.

B

But doesn't the height or the altitude have to be perpendicular to the base? Thereby creating a right triangle - ADB = 90=CDB??

The height does have to be perpendicular to the base--but BD isn't necessarily the height. If ADB and CDB aren't right angles, then you would have to drop a perpendicular from B to an extension of the line AC for the height. (Note that the height of a triangle isn't necessarily contained inside the triangle.)

Another way of looking at it: given statement 1, it's possible that BD is perpendicular to AC, in which case ABC is isosceles. But it's also possible that ABC looks just like the drawing, with AD = CD, but ADB < CDB, so that AB < CB.
Re: Isosceles Triangle   [#permalink] 28 Nov 2007, 00:46
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