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Is side BC parallel to DE ?

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by goelmohit2002 » Sat Sep 12, 2009 10:13 pm
In triangle ABC, D is a point on side AB. E is a point on side AC. Is DE parallel to BC ?

a) D is midpoint of AB.
b) E is midpoint of AC.

OA after some discussion.
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by Nermal » Sun Sep 13, 2009 2:22 am
IMO C
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by goelmohit2002 » Sun Sep 13, 2009 3:21 am
Nermal wrote:IMO C
Can you please give your reasoning for the same ?
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by scoobydooby » Sun Sep 13, 2009 9:44 am
comes from the mid-point theorem of triangles: a line joining the midpoints of a triangle is parallel to the third side and is half the length of the 3rd side.
Last edited by scoobydooby on Sun Sep 13, 2009 10:52 am, edited 2 times in total.
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by Nermal » Sun Sep 13, 2009 10:37 am
scoobydooby wrote:we need niether of the options. it follows from the stem itself that DE is parallel to BC and is equal to 1/2 BC

comes from the mid-point theorem of triangles: a line joining the midpoints of a triangle is parallel to the third side and is half the length of the 3rd side.
Of course we need both of the statements since the question stem only tells us that the points E and D lie somewhere on the respective sides of the triangle.
The statements tell us that both of the points are the midpoints of the sides. Without that information we cannot solve this problem at all!
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by scoobydooby » Sun Sep 13, 2009 10:47 am
oops, you are right nermal. i was careless enough to read "points' as "mid-points'. my bad!
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by goelmohit2002 » Sun Sep 13, 2009 10:07 pm
Thanks. But can you pkease prove this theorem mathematically....i tried proving the same using similar triangles concept...but was not able to do so....
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Re: Is side BC parallel to DE ?

by euwe » Mon Sep 14, 2009 12:38 am
goelmohit2002 wrote:In triangle ABC, D is a point on side AB. E is a point on side AC. Is DE parallel to BC ?

a) D is midpoint of AB.
b) E is midpoint of AC.

OA after some discussion.
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by Nermal » Mon Sep 14, 2009 2:42 am
I thought my my way through this problem and looked for a mathematical reasoning afterwards.
What I found was the so called RAR-rule of similar triangles.
It says that triangles are similar (similar meaning having the same measures of angles but different side lenghts) when one angle is the same (here: angle at vertex A) and the sides that contain this angle are in the same ratio (here: AE=(1/2)AC and AD=(1/2)AB).
Therefore the other angles are the same as well and this can only be true if the sides CB and DE are parallel.

Look here: https://www.onlinemathlearning.com/simil ... ngles.html
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by goelmohit2002 » Mon Sep 14, 2009 11:42 am
Nermal wrote:I thought my my way through this problem and looked for a mathematical reasoning afterwards.
What I found was the so called RAR-rule of similar triangles.
It says that triangles are similar (similar meaning having the same measures of angles but different side lenghts) when one angle is the same (here: angle at vertex A) and the sides that contain this angle are in the same ratio (here: AE=(1/2)AC and AD=(1/2)AB).
Therefore the other angles are the same as well and this can only be true if the sides CB and DE are parallel.

Look here: https://www.onlinemathlearning.com/simil ... ngles.html
Awesome Nermal !!!

Got the same....actually since Midpoint theorem is not in the course of GMAT(nowhere OG talks about the same)...so I was hesitant to use the same directly and wanted to learn the approach that OG suggests.
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by goelmohit2002 » Mon Sep 14, 2009 11:43 am
Thanks Guys for the discussion....OA = C.
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by GambitOS » Mon Sep 14, 2009 11:33 pm
:?:
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by Nermal » Tue Sep 15, 2009 1:58 am
@GambitOS
Your point E is definitely not the midpoint of AC.
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by goelmohit2002 » Tue Sep 15, 2009 3:23 am
GambitOS wrote:Really? :)
Hi Gambit,

From Naked Eye I am not able to find out whether it is midpoint or not...can you please tell how to check the same in the figure that u drew ?

Any software that I can test ur figure ?

Thanks
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by GambitOS » Tue Sep 15, 2009 3:27 am
:oops: Sorry guys, I recheck my figure. It's really parallel! :)
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