AliasgharA585 wrote:Thanks Ron for the explanation...

you're welcome.

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- RonPurewal
- Students
**Posts:**19746**Joined:**Tue Aug 14, 2007 8:23 am

AliasgharA585 wrote:Thanks Ron for the explanation...

you're welcome.

- EnriqueR905
- Students
**Posts:**34**Joined:**Sun Mar 13, 2016 1:41 pm

Hello, i still do not understand why it is not A. It seeems like since if we look at the image from the question we can clearly observe that we have a right angle and then statement 1 says that line ac is perpendicular,doesn´t that mean that the 90 degrees angle is cut in half and because we have a perpendicular line,we have then a 45-45-90 triangle which would make statement 1 sufficient by itself?.

Could someone help me clarify this?

Thank You.

Could someone help me clarify this?

Thank You.

- tim
- ManhattanGMAT Staff
**Posts:**5665**Joined:**Tue Sep 11, 2007 9:08 am**Location:**Southwest Airlines, seat 21C

We still don't have a diagram. We will be glad to help you once someone posts the original diagram from the problem.

Tim Sanders

Manhattan GMAT Instructor

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- EnriqueR905
- Students
**Posts:**34**Joined:**Sun Mar 13, 2016 1:41 pm

Hello, i am sorry i could not upload the image directly,but here it is: http://gmatclub.com/forum/download/file.php?id=13493

- tim
- ManhattanGMAT Staff
**Posts:**5665**Joined:**Tue Sep 11, 2007 9:08 am**Location:**Southwest Airlines, seat 21C

Thanks. Your mistake is assuming that angle A is always bisected by altitude AC. That occurs if and only if the large triangle is isosceles, so you used an assumption that the triangle was isosceles to prove that it is isosceles.

Tim Sanders

Manhattan GMAT Instructor

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- ZadiaH581
- Students
**Posts:**9**Joined:**Wed Oct 14, 2015 8:01 am

Hello,

I know this is an older post, but I was wondering since Statement 1 says that AC is perpendicular to BD, don't we automatically know that that BC and DC are the same length, since perpendicular bisectors are known to divide a line segment exactly in half? If that is the case, should A not be sufficient to know that the two triangles are similar? Also, does AC divide angle BAD also in half? Or does this rule not happy to perpendicular lines in triangles?

Thank you for your help in advance.

I know this is an older post, but I was wondering since Statement 1 says that AC is perpendicular to BD, don't we automatically know that that BC and DC are the same length, since perpendicular bisectors are known to divide a line segment exactly in half? If that is the case, should A not be sufficient to know that the two triangles are similar? Also, does AC divide angle BAD also in half? Or does this rule not happy to perpendicular lines in triangles?

Thank you for your help in advance.

- RonPurewal
- Students
**Posts:**19746**Joined:**Tue Aug 14, 2007 8:23 am

sure, but, statement 1 doesn't say "perpendicular bisector"... it just says "perpendicular".

so, there's no reason why C would have to be the midpoint of BD.

so, there's no reason why C would have to be the midpoint of BD.

- LukeT323
- Students
**Posts:**1**Joined:**Sat Aug 13, 2016 1:01 am

tim wrote:Thanks. Your mistake is assuming that angle A is always bisected by altitude AC. That occurs if and only if the large triangle is isosceles, so you used an assumption that the triangle was isosceles to prove that it is isosceles.

Shouldn't I assume that AC splits angle BAD into two 45 degree angles?

- RonPurewal
- Students
**Posts:**19746**Joined:**Tue Aug 14, 2007 8:23 am

LukeT323 wrote:Shouldn't I assume that AC splits angle BAD into two 45 degree angles?

...most definitely not. if the right angle is formed by 2 sides of significantly different lengths, then those angles won't be anywhere even close to 45° each.

...as in this figure

http://www.mathplanet.com/Oldsite/media ... 99x300.jpg

it should be absolutely clear that angles ABD and DBC in that figure are nowhere close to 45°. (angle ABD is much bigger than 45°, and angle DBC is much smaller than 45°.)