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parallel?

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) ∠BAD = ∠ADC


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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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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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by arora007 » Sun Jan 16, 2011 1:41 am
realize my folly!! thanks rahul.
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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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by Ramit88 » Sun Jan 16, 2011 10:27 am
plz explain how AB ll to CD ?
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by anshumishra » Sun Jan 16, 2011 10:59 am
Ramit88 wrote:plz explain how AB ll to CD ?
∠BAD = ∠ADC
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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
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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 :)
Attachments
Parallel.png
Last edited by shovan85 on Sun Jan 16, 2011 11:10 am, edited 1 time in total.
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by Ramit88 » Sun Jan 16, 2011 11:10 am
how statement 1 is suff to say that AB ll CD
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by anshumishra » Sun Jan 16, 2011 11:21 am
Ramit88 wrote:how statement 1 is suff to say that AB ll CD
∠AEB =∠CED
∠ECD =∠ EDC
AE*ED = BE*CE (theorem of intersecting secants) : [url] https://www.mathopenref.com/chordsintersecting.html
Since, CE = DE => AE = BE
So, ∠EAB =∠ EBA = ∠ECD =∠ EDC

Hence, AB ll CD
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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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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
∠AEB =∠CED Why?
∠ECD =∠ 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 ∠ECD =∠ EDC i.e. CE=DE then AE=BE i.e ∠ECD =∠ EDC but dat does not mean that all the four below are equal)
So, ∠EAB =∠ EBA = ∠ECD =∠ EDC

Hence, AB ll CD
Some doubts.Apology if these are too silly.
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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
∠AEB =∠CED Why?
∠ECD =∠ 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 ∠ECD =∠ EDC i.e. CE=DE then AE=BE i.e ∠ECD =∠ EDC but dat does not mean that all the four below are equal)
So, ∠EAB =∠ EBA = ∠ECD =∠ 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
∠AEB = ∠CED = z
∠ECD = ∠EDC = x
∠EAB = ∠EBA = y
So, x+x+ z = 180 = y + y+ z
=> 2x = 2y => x=y
So, ∠EAB =∠ EBA = ∠ECD =∠ EDC

Hope that clears your doubt.
Thanks
Anshu

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