parallel?

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Deepthi Subbu: Master | Next Rank: 500 Posts; Posts: 298; Joined: Tue Feb 16, 2010 1:09 amFollowed by: 1 member; Thanked: 2 times

by Deepthi Subbu » Sun Jan 16, 2011 1:00 am

Segments AD and BC intersect at point E. E lies on segment FG. Are the lines FG and AB parallel to CD?

(1) Triangle CED is isosceles with base CD.

(2) âˆ BAD = âˆ ADC

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arora007: Community Manager; Posts: 1048; Joined: Mon Aug 17, 2009 3:26 amFollowed by: 27 members; Location: India; Thanked: 51 times; GMAT Score:670

by arora007 » Sun Jan 16, 2011 1:29 am

B is pretty obvious , SUFFICIENT.

well as for A, " Triangle CED is isosceles with base CD"

assuming that it is saying CE=ED (thats what i make out of this sentence) then because of similarity of oppsite triangles in a circle should also be SUFFFICIENT,

so my take would be D.

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Sun Jan 16, 2011 1:36 am

The problem is asking "Are the lines FG and AB parallel to CD?"

If it was only AB, then each of the statements alone is sufficient.
But as it is 'FG and AB', none of the statements provide any information that can allow us to relate FG with AB and CD. Hence we cannot specify any single line FG as there are infinite possible lines that would pass through E.

Hence, together the statements are not sufficient to answer the problem.

The correct answer is E.

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arora007: Community Manager; Posts: 1048; Joined: Mon Aug 17, 2009 3:26 amFollowed by: 27 members; Location: India; Thanked: 51 times; GMAT Score:670

by arora007 » Sun Jan 16, 2011 1:41 am

realize my folly!! thanks rahul.

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ankur.agrawal: Master | Next Rank: 500 Posts; Posts: 261; Joined: Wed Mar 31, 2010 8:37 pmFollowed by: 3 members; Location: Varanasi; Thanked: 11 times

by ankur.agrawal » Sun Jan 16, 2011 3:31 am

What is the rule for two lines to be PArallel. Can sumbody plz explain?

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Ramit88: Senior | Next Rank: 100 Posts; Posts: 67; Joined: Fri Nov 26, 2010 8:25 am; Thanked: 3 times

by Ramit88 » Sun Jan 16, 2011 10:27 am

plz explain how AB ll to CD ?

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Sun Jan 16, 2011 10:59 am

Ramit88 wrote:plz explain how AB ll to CD ?

âˆ BAD = âˆ ADC

Thanks
Anshu

(Every mistake is a lesson learned )

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Sun Jan 16, 2011 11:01 am

ankur.agrawal wrote:What is the rule for two lines to be PArallel. Can sumbody plz explain?

https://library.thinkquest.org/20991/geo/parallel.html

Thanks
Anshu

(Every mistake is a lesson learned )

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shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

by shovan85 » Sun Jan 16, 2011 11:03 am

See the below picture that shows the properties of line to be parallel.

When you know:

1. Angle AFG = Angle FGD,
2. Angle CGF = Angle BFG,
3. Angle EFB = Angle EGD,
4. Angle EFA = Angle EGC,
5. Angle AFG = Angle CGH,
6. Angle BFG = Angle DGH,
7. Angle AFG + Angle EGC = 180,
8. Angle BFG + Angle FGD = 180,

Any of the above is given to you then you are definitely dealing with a pair of parallel lines. And you can also derive couple of more formulas to know the parallelism

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Ramit88: Senior | Next Rank: 100 Posts; Posts: 67; Joined: Fri Nov 26, 2010 8:25 am; Thanked: 3 times

by Ramit88 » Sun Jan 16, 2011 11:10 am

how statement 1 is suff to say that AB ll CD

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Sun Jan 16, 2011 11:21 am

Ramit88 wrote:how statement 1 is suff to say that AB ll CD

âˆ AEB =âˆ CED
âˆ ECD =âˆ EDC
AE*ED = BE*CE (theorem of intersecting secants) : [url] https://www.mathopenref.com/chordsintersecting.html
Since, CE = DE => AE = BE
So, âˆ EAB =âˆ EBA = âˆ ECD =âˆ EDC

Hence, AB ll CD

Thanks
Anshu

(Every mistake is a lesson learned )

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ankur.agrawal: Master | Next Rank: 500 Posts; Posts: 261; Joined: Wed Mar 31, 2010 8:37 pmFollowed by: 3 members; Location: Varanasi; Thanked: 11 times

by ankur.agrawal » Sun Jan 16, 2011 10:48 pm

anshumishra wrote:
ankur.agrawal wrote:What is the rule for two lines to be PArallel. Can sumbody plz explain?
https://library.thinkquest.org/20991/geo/parallel.html

Thanks buddy. Dat was a good link.

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ankur.agrawal: Master | Next Rank: 500 Posts; Posts: 261; Joined: Wed Mar 31, 2010 8:37 pmFollowed by: 3 members; Location: Varanasi; Thanked: 11 times

by ankur.agrawal » Sun Jan 16, 2011 11:21 pm

anshumishra wrote:
Ramit88 wrote:how statement 1 is suff to say that AB ll CD
âˆ AEB =âˆ CED Why?
âˆ ECD =âˆ EDC (we know it is an insocles triangle with base CD but does that means EC=ED. It can be CE=CD or even CD=ED.
AE*ED = BE*CE (theorem of intersecting secants) : [url] https://www.mathopenref.com/chordsintersecting.html
Since, CE = DE => AE = BE ( if we even assume âˆ ECD =âˆ EDC i.e. CE=DE then AE=BE i.e âˆ ECD =âˆ EDC but dat does not mean that all the four below are equal)
So, âˆ EAB =âˆ EBA = âˆ ECD =âˆ EDC

Hence, AB ll CD

Some doubts.Apology if these are too silly.

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Mon Jan 17, 2011 5:53 am

ankur.agrawal wrote:
anshumishra wrote:
Ramit88 wrote:how statement 1 is suff to say that AB ll CD
âˆ AEB =âˆ CED Why?
âˆ ECD =âˆ EDC (we know it is an insocles triangle with base CD but does that means EC=ED. It can be CE=CD or even CD=ED.
AE*ED = BE*CE (theorem of intersecting secants) : [url] https://www.mathopenref.com/chordsintersecting.html
Since, CE = DE => AE = BE ( if we even assume âˆ ECD =âˆ EDC i.e. CE=DE then AE=BE i.e âˆ ECD =âˆ EDC but dat does not mean that all the four below are equal)
So, âˆ EAB =âˆ EBA = âˆ ECD =âˆ EDC

Hence, AB ll CD
Some doubts.Apology if these are too silly.

No need to apology, sometimes I may miss something as well (hopefully not in this case). Keep the doubts coming

As derived above : CE = DE => AE = BE
Look at the triangles AEB and CED
âˆ AEB = âˆ CED = z
âˆ ECD = âˆ EDC = x
âˆ EAB = âˆ EBA = y
So, x+x+ z = 180 = y + y+ z
=> 2x = 2y => x=y
So, âˆ EAB =âˆ EBA = âˆ ECD =âˆ EDC

Hope that clears your doubt.

Thanks
Anshu

(Every mistake is a lesson learned )

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