circle geo question (1 Viewer)

[Grey Council]

AB and CD are two parallel chords. The tangent to the circle at D meets AB produced at T. The other tangent from T touches the circle at E. EC cuts AB at F.

a) Prove E, F, D and T are concyclic.
b) Prove FC = FD
c) Hence prove that AF = FB
d) Prove that OFET and OFTD are cyclic quadrilaterals, where O is the centre of the circle. (It may be helpful to draw a new diagram for the sake of clarity).

wtf, i just did two other question 8 circle geo questions before this, and got really proud ( ) and then hit this. I've been staring at that question for literally 40 minutes. I think i'll leave it for a while.

btw, if its really simple, don't give me solution or any hints. Just say its simple.
If its hard, though, just give a hint.

thanks

btw, anyone heard of the 3 minute rule for circle geo? ^__^

 

[nike33]

ive got 2 answers, one im pretty sure is right, the other (considerably shorter, and not 100%)

let intersectionpt of FT and ED = B

angle FBE = angle DBT vert opp..

angle DCE = alpha = angle TFE (correspoding in parrallel lines)
angle DCE = alpha = angle TDE (angle at tangent / chord is equal to angle in alternate segment)

angle TDB = angle TFE
angle FBE = angle DBT
angle BEF = angle BTD

draw an assumend circle around FETD now it IS a circle as all F E T D lie on the cirlce according to angles in the alternate segment are equal...hence concyclic...i cant expalin it that well

 

[:: ck ::]

for ur first attempt... i dunno if ur allowed to say that EBD is a straight line... as the question doesnt state it ... or are u? ....

and the second one... ermm ur starting with the assumption that EFDT is already concyclic... which doesnt really make sense... unless u can elaborate?

 

[Affinity]

It's simple..
Whenever you get circle geometry questions, there's virtually only 2 things you could do.. and it's usually the first: angle bashing.

 

[CM_Tutor]

Nike33's first answer does work - with one change. The question defines a point B, at one end of one of the parallel chords. If the statement "let intersectionpt of FT and ED = B" is meant to refer to this B, then the proof is certainly flawed, because (as Ryan.cck points out), EBD isn't a straight line. However, I think that nike33 has forgotten that B was already defined. So, change the proof to read "let intersectionpt of FT and ED = X", and then change every B in his proof to an X, and the proof is OK.

If the statement "draw an assumed circle ..." is supposed to be another proof, then I'm not following it, and I would share ryan.cck's concern that it seems to start with an assumption that the result is true, which is not allowed.

Grey Council, there is a four line proof for this result. Ask yourself this question: What ways are there to prove that four points are concyclic?

 

[Grey Council]

lol, fair enough.

Will try again tommorow. I've been doing maths since 7am. humph, i really should study my other subjects. hehe, relaxing now, then will go do some 4u eng. ^__^ 4u eng is such a bludge.

was typing same time as CM_Tutor, apparantly. hehe, I'll try tommorow Tutor. This is the first part of the question, so it SHOULD be easy. hehe, i dunno, had a block. thanks though

 

[nike33]

ahh yes i should have called it something other than B hehe sry...i didnt mean the B on chord AB...


and the "assume draw a circle.." thingy was just a poor attempt at explaining what i was trying to say, not another proof

 

[MyLuv]

Heres an another short and easy way to do this:
ang TDE= ang DCE ( arc DE)
but ang FED+EDC+ECD=180 --> FED+TDE+EDC=180 ie FED+TDC=180

and FTD+TDC=180 (AB//CD)

--> FTD=FED
--->F,T,D,E concyclic

 

[Euler]

Originally posted by Affinity
It's simple..
Whenever you get circle geometry questions, there's virtually only 2 things you could do.. and it's usually the first: angle bashing.

that's generally the way to go...

i would call it "angle chasing"....bashing is for cases, i.e. case bashing

 

[CM_Tutor]

If you're going to do angle chasing, do it quickly on a diagram first to see where you can get. Then look to see what is actually helpful. Otherwise, you can waste a lot of time.

PS: I only advocate angle chasing after giving some thought to your goal - otherwise you can chase angles that couldn't possibly help you.

 

[Grey Council]

lol, so true CM_Tutor, thats what I was doing. but the problem is, you can attempt to prove its a cyclic quad so many different ways, esp in this question, there are SSSOOO many angles. AND its question 8, so i was like, this is gonna be hard. lol, turned out rather simple, alternate segment theorem and corresponding angles in parallel lines. ^__^
but I found the four line result WITHOUT looking at any of the answers people posted up.
my friends, i didn't ask for a solution if it was an easy question. thanks anyway though.

anyway, that was just the first part of the question. Now i'm stuck on the third and fourth bits:
Prove AF = FB
will post up last part after getting this. I wanna have a go at last bit myself, after proving third bit.

 

[CM_Tutor]

Grey Council, it's good that you found the simple solution to this problem, and I agree that there are a variety of ways to establish that four points are concyclic - this question makes use of the converse of the angles of the same arc theorem, and in my experience it is this approach that people have most trouble seeing.

PS: This is a fairly common question - you will see it on a number of trial papers, so it may be worth you posting up the whole question for everyone to try once you are ready to do so.

 

[Grey Council]

but, but, how do i do the third bit?!? lol, will post up whole question. in fact, have edited first part and have posted up full question.

 

[nike33]

if u still need help...ill give some hints


b) easy just show iso triangle
c) FC = FD (just proved)
AC = BD (as angle AFC = angle DFB..equal angles subtend equal chords etc...) hence show triangles are congruent..then FA = FB ie corresponding sides in cong. triangle
d) should be fairly simple if u have the angles labelled..finding angles in alternate segment are equal proves this easily, among various other proofs

 

[CM_Tutor]

As I said above, this question has popped up on a number of previous trials, including Newington 1993, Q 6(b) and Knox 2001, Q 8(a). In those cases, you are led through the question differently, so here's the Newington 1993 version:

PQ and CD are parallel chords of a circle. The tangent at D cuts PQ produced at T. B is the point of contact of the other tangent from T, and BC meets PQ at R.

(a) Draw a diagram to illustrate this information.

(b) Prove that angle BDT = angle BRT, and hence state why B, T, D and R are concyclic points.

(c) Prove that angle BRT = angle DRT.

(d) Show that triangle RCD is isosceles.

(e) Prove that triangles PRC and QRD are congruent.

(f) Hence, or otherwise, prove that BC bisects PQ.

 

[Grey Council]

Originally posted by nike33
if u still need help...ill give some hints
b) easy just show iso triangle
c) FC = FD (just proved)
AC = BD (as angle AFC = angle DFB..equal angles subtend equal chords etc...) hence show triangles are congruent..then FA = FB ie corresponding sides in cong. triangle

hehe, was up to c anyway, and did that last night. Thanks for the hints.

d) should be fairly simple if u have the angles labelled..finding angles in alternate segment are equal proves this easily, among various other proofs

Now up to this. hrm, working on it.

Thanks CM_Tutor. Your question doesn't have the last part of my one, does it?
I got my one from, and I quote:
4U Paper 2 - Prior Education Australia
The Ultimate Trial HSC Mathematics Revision course 1994

What really freaked me out was when I found out that this question is only question 3. Not a question 7/8. :-\

 

[CM_Tutor]

Originally posted by Grey Council
Thanks CM_Tutor. Your question doesn't have the last part of my one, does it?
I got my one from, and I quote:
4U Paper 2 - Prior Education Australia
The Ultimate Trial HSC Mathematics Revision course 1994

What really freaked me out was when I found out that this question is only question 3. Not a question 7/8. :-\

Prior have some interesting ideas on what is, and is not, reasonable, IMO.

As for the last part, I haven't looked at it - part of the reason for posting the other sources is that I have come across this question many times before, but its goal is usually to show the bisection of the chord. Will have a look later on...

 

[Grey Council]

Originally posted by CM_Tutor
Prior have some interesting ideas on what is, and is not, reasonable, IMO.

As for the last part, I haven't looked at it - part of the reason for posting the other sources is that I have come across this question many times before, but its goal is usually to show the bisection of the chord. Will have a look later on...

hehe, true true. Although Prior is the most successful coaching college in Sydney.

But that has mostly got to do with the 6ish hours of maths they make every student do on top of school work. ^_^ ANYONE works that much, they'll be good at maths.

And I still can't do last part. :-\
I can't see any alternate segments. help?

 

[:: ck ::]

lolz i go prior i wudn call myself gud at maths... lol

id say 1/3 of the students there actually try and do well.. the others well.. they are just wasting money [A LOT OF MONEY] ...

 

[CM_Tutor]

Originally posted by Grey Council
hehe, true true. Although Prior is the most successful coaching college in Sydney.

But that has mostly got to do with the 6ish hours of maths they make every student do on top of school work. ^_^ ANYONE works that much, they'll be good at maths.

No Comment.

And I still can't do last part. :-\
I can't see any alternate segments. help?

I'll have a look and get back to you.