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.

realize my folly!! thanks rahul.

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

plz explain how AB ll to CD ?

Ramit88 wrote:plz explain how AB ll to CD ?

∠BAD = ∠ADC

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

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

how statement 1 is suff to say that AB ll CD

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

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.

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.

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.