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fashionista

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hiya!!! if theres anyone out there who can ease my pain please do so...i was cruisin thru sum kwestions wen i got to this one and my mind jus wiped itself clean. bluddy question 27!!!! neways heere it is
27. AB and AC are equal chords of a circle. AD and BE are parallel chords through A and B respectively. Prove that AE is parallel to CD. ok..so if my computer will cooperate u shud see a lil diagram of my kwestion below..if not im really sorry :D
:vcross: :mad: :chainsaw: k..so my computaa aint gonna do it for me but pleasssse can sum1 put my mind at ease so i can sleep easy tonite??????
thanx heaps!!
 

iambored

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Originally posted by fashionista
i got that angle AEB = angle EAD.cuz they're alternate angles but thats it
yeah i think you use that, and then that CDA = EAD because they are alternate

umm, and you may be ablt to use similar triangles?
i really don't know, i can't seem to work it out
 

Rahul

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Originally posted by iambored
that CDA = EAD because they are alternate
?
i was getting CDA and EAD were supplementary. or ACD = CAE, because they are alternate. :confused:

meh, maybe i drew the diagram wrong...its too late for maths i say :p
sorry fashionista. :)
 

Lainee

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I think I've figured out something:

1. Draw in DE. For AE to be parallel to CD, DE=AC
2. Because AC=AD (given), you just need to use similar triangles (I'm not going to go through this, as it's fairly straight fwd) to prove that DE=AB=AC
 

iambored

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Originally posted by Rahul
?
i was getting CDA and EAD were supplementary. or ACD = CAE, because they are alternate. :confused:

meh, maybe i drew the diagram wrong...its too late for maths i say :p
sorry fashionista. :)
by the way, i was saying things which show that they are alternate. so if you can prove that bea = ead = adc then you can get alternate angles, as they are qeual
 

kimmeh

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Originally posted by Rahul
meh, maybe i drew the diagram wrong...its too late for maths i say :p
sorry fashionista. :)
lol i rekon.. though i have seen this question before somewhere..heres a starter :p
umm well i let angle CBA = beta
.: < AEC = beta (same seg)
.: < ACB = beta (base angles of isos triangle)
.: < AEB = beta (same seg)
angle CBE = alpha
.: < CAE = alpha (same seg)
now all you have to do is prove < DCA equals alpha...somehow :confused:
 
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fashionista

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rahul and iambored...i did wut u guys did too..but then i figured u cant assume that these angles r alternate because u r trying to prove that AE is parallel to CD first. so for them to be alternate ud have to prove that.
 

Lainee

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fashionista - have you tried what I suggested in my post? I'm not seeing what Rahul and iambored is heading towards... but I also don't see anything wrong with mine at the moment (although I may soon be going ahhhhh and run out embarassed) :p
 

ND

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Let T be a point on AD extended (so that D is in the middle)
EAD=BEA (alternate)
AEB=ACB (angles in same segment are equal)
ACB=ABC (since AB=AC, /\ABC is isosceles, and base angles are equal)
.'.AEB=ABC
.'.ABC=EAD
ABC=CDT (ext angle is equal to interior opp angle in a cyclic quad)
.'. EAD=CDT
.'. AE||CD (adjacent)

First bit of maths since the 3u exam...
 

fashionista

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oooohhhh!! thanx evereebodee!!...is there any possible way of doing this kwestion without the extension of AD?
sorry rahul...i tried to post a diagram but it wudnt work
 

fashionista

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Lainee i tried ur method too!! umm im not sure it totally made sense to me but then again maybe thats just my fault...i kinda got lost wen u sed the ' for AE to be parallel to CD, DE=AC'..cuz it sounded a bit assumptiony to my hed. thanx for ur help tho!!
 

Lainee

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LOL, goes to show my explanation skills if you got lost at my first step. :p

yeah it is an assumption but a fairly obvious one. If you have two lines, x and y, and a and b are two pts on x and c and d are two pts on y - then for x and y to be parallel, ac=bd. It's like saying that if two lines are to be parallel then all the pts on one has to be equal distance to all the points on the other. ;)
 

ND

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Hmmm, Lainee that's not necessarily true; it only works if the lines are parallel. But i don't see how that relates to this question. Also, AB=AC is given, not AC=AD.
 

Lainee

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Originally posted by ND
it only works if the lines are parallel.
Isn't that what we're typing to prove? that AE and DC are parallel?

Originally posted by ND
But i don't see how that relates to this question. Also, AB=AC is given, not AC=AD.
Well, if you scroll up ^ I posted my idea to it. Sorry about that, yes, AC=AB, scribing error on my part. The method still works though... except that now on second though, proving that the triangles are similar may be harder than I initially thought.
 

ND

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Originally posted by Lainee
Isn't that what we're typing to prove? that AE and DC are parallel?
What you're describing is a parallelogram, but there is no parallelogram in the question. I mean AC always intersects DE.
 

Lainee

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Originally posted by ND
What you're describing is a parallelogram, but there is no parallelogram in the question. I mean AC always intersects DE.
Hmm... in my diagram ACDE shows a rhombus if AE and DC are parallel. AC goes no where near to intersecting DE on my diagram. :chainsaw: Any chance of you scanning in your diagram so I can see what you guys are seeing? :p I check and double check mine but there's nothing wrong with it!

Is it possible that sometimes given inadequate information two ppl may end up with different diagrams... fashionista - did you get this question from a textbook?
 

fashionista

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yes i did...fitzpatrick 3u one... on my diagram..which i tried to post ACDE makes a trapezium and AC does by no means intersect DE. does any one noe how to post a piccie??
 

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