circle geo question (2 Viewers)

[CM_Tutor]

I've thought about (d). Both parts are really easy - about 2-3 lines for each - once you realise the trick / correct approach here (and I did a fair amount of angle chasing before I saw it! ).

Hint #1: What is angle OET?

Hint #2: What is the relationship between the lines AB and OF and why?

Comment: I think this would've been a better question as "Show that OT is the diameter of a cricle through E, F, O, D and T".

 

[Giant Lobster]

i did parts a - c cbf drawing a new diagram for d) haha

 

[Grey Council]

Originally posted by CM_Tutor
Both parts are really easy - about 2-3 lines for each - once you realise the trick / correct approach here (and I did a fair amount of angle chasing before I saw it! ).

hang on a second . . . Thats kinda dodgy! lol, every question is REALLY easy once you see how to do it. Once you realise the trick / correct approach. Thing is, you hafta do a lot of angle chasing for the hard ones. ^_____^

Hint #1: What is angle OET?

90 degrees

Hint #2: What is the relationship between the lines AB and OF and why?

OF bisects the line AB. Therefore OF is perpendicular to AB.

how does it help me?
hang on, doing now

 

[Grey Council]

hang on a second...
You gave it away. lol, i think.

Originally posted by CM_Tutor
Comment: I think this would've been a better question as "Show that OT is the diameter of a cricle through E, F, O, D and T".

so, is it cause we know that EFDT is concyclic, and OT subtends a 90 degree angle to F and to E, therefore OT diameter. Therefore it must also subtend a 90 degree angle to D. which means ODT is 90. Which means ODT and OFT are concyclic points (opposing sides are supplementary).

and OED = OFD (90 - alpha, reason listed in previous post)

lol, i figured out the second one without even realising.

 

[Grey Council]

allright, next question:
Two unequal circles meet at A and B. The smaller circle passes through O, the centre of the larger circle. The tangent to the smaller circle at B meets the larger circle again at C. Prove that triangle ABC is isosceles.

Am doing now.

EDIT: I can't say that AC is a tangent, and stops at C. Therefore a line from C to B is another tangent, therefore BC = AC, can I? we don't know that CB is a tangent to teh smaller circle.
hrm, nice question, i haven't a clue where to start. shh, i'll keep trying

 

[nike33]

Let angle (BCA) = @

(BOA) = 2@
(ABX) = 2@ (where x is on cb subtended)

.:. (BCA) + (CAB) = 2@ (exterori angle of a tri = sum of opp interor angles)

@ + (CAB) = 2@
(CAB) = @
(BCA) = @ .:. ABC iso

 

[Grey Council]

can you hide your answers? damn it, some people are trying to do them. :-\

and why is:
(ABX) = 2@ ? hrm

EDIT
ARGH! I said BAO is equal to ABX. :-@
humph, well done, nike33. Although I would have got that, if it wasn't for some silly mistake I made. ah well, close isn't good enough, aye?

well done

 

[CM_Tutor]

Originally posted by Grey Council
hang on a second . . . Thats kinda dodgy! lol, every question is REALLY easy once you see how to do it. Once you realise the trick / correct approach. Thing is, you hafta do a lot of angle chasing for the hard ones. ^_____^

No, I didn't mean it that way. What I meant was there is a really easy, 2 - 3 line approach. If you see it quickly, you won't have a problem with the question. In this case, I didn't see it quickly, and so wasted time with angle chasing. What I should have done was ask myself what the centre O told me, which is that, since OF bisects AB, OF must be perpendicular to AB, and that angles OET and ODT are right angles.

It then immediately follows that OFET is a cyclic quad (angles on the same arc OT are equal) and OFDT is a cyclic quad (opposite angle OFT and ODT are supplementary.

Since OT is the diameter of both circles OFET and OFDT, they must be the same circle, and the result I suggested follows.

I am not saying that all questions are easy - far from it - but this one is as soon as you look at it the right way. A lot of the end of 4u papers is like this, and spotting short cuts is a necessary skill to develop.

 

[Grey Council]

lol, fair enough. I take what I said back.
I was joking anyway.

 

[nike33]

how do you hide answers? anyway i only posted it cause i thought u neded help with it

 

[:: ck ::]

shove it in a html tag like < answer here >

 

[Grey Council]

lol, I just posted that question up, I would have found my mistake soonish. ah well, like I said, well done.

And to hide solution, do as Ryan does. ^_^

<hello, i'm testing this for the first time too lol>

quote my text, above here should be revealed a hidden line.

 

[Grey Council]

Next question:
Let ABCD be a cyclic quad. E is a point on AC such that ABE = DBC

i) Show that triangle ABE ||| triangle DBC
ii) Show that triangle ABD ||| triangle EBC
iii) Hence show that AB.DC + AD.BC = AC.DB

heh, am stuck on last bit. :-\

 

[nike33]

have u got answers? heres my attempt

AC = AE + EC (3)

AE / DC = AB / DB (from parts i and ii)
AE = (AB.DC / DB) (1)
EB / EC = AD / AB (couldnt find a better way of doing it so i proved those are similar)
AD.EB = AB.EC
EC = (AD.EB) / AB (2)

sub 1 and 2 into 3

AC = (AB.DC / DB) + (AD.EB) / AB
AC.DB = AB.DC + DB.AD.EB / AB
now DB.EB / AB = BC

sub that in and you get AC.DB = AB.DC + AD.BC

 

[Grey Council]

huh, if I had answers, I wouldn't be asking.

EDIT:
My sincere apologies. I had posted up the question a tad bit wrong.

Have changed it.
Should be
ii) Show that triangle ABD ||| triangle EBC
not
ii) Show that triangle ABD ||| triangle EBD

apologies

 

[riVa0o]

iii)
we kno triangle ABD||| triangle EBC
so AB/EB = AD/EC = DB/BC ...(1) (don't make me write why ><)
we also kno that triangle ABE ||| triangle DBC
so AB/DB = BE/BC = AE/DC ...(2)

so from (1) AD.BC = DB.EC
and from (2) AB.DC = DB.AE

adding the two equations

AB.DC + AD.BC = DB.EC + DB.AE = DB(EC+AE) = DB.AC
and therefore,
AB.DC + AD.BC = DB.AC

hope ur happy now dude >< gah forcin me to post

 

[Grey Council]

nice, Riva, nice.

Will post up more questions tommorow, when I get stuck on one. heheh

 

[nike33]

haha thats why i couldnt use ii and had to use another triangle in my answer