annoying circle geo (1 Viewer)

[mojako]

Hi... looks like I have a question to ask
It's from the syllabus, page 83.

BF, FE, AE and BD are 4 straight lines as in the attachment.
AE meets BD at H.
Circles are drawn through the vertices of the four triangles ABH, HDE, FBD and FAE.
Prove that the four circles have a common point.

Thank you ^_^

 

[Xayma]

Hmm at a quick look prove The circles AHB and DHE intersect at H and another spot, let this point be Q. Prove that FDQB are cyclic and FAQE are cyclic would do it. As to how to do that, Im currently going through theorems.

 

[mojako]

ill try my best to do it

but Xayma. r u sure u r not doing ext2??
coz.. its just strange looking at how much u write in the ext2 forum

EDIT: doesnt look like I can do it

 

[Archman]

isnt there a initial condition of A,B,E,D being concyclic?

 

[mojako]

isnt there a initial condition of A,B,E,D being concyclic?

umm..
the syllabus doesnt mention it
but maybe they left it out..
coz its just an illustration of how hard the questions should be

But I tried making random lines.. doesnt check if A B E D concyclic.. using Ms Paint.. and the points of intersections come pretty close
and my ABED doesnt seem to be concyclic

 

[mojako]

because the result is not completely true if they are not
But, if they are, then the result is true.
so try to do the question taking that abed is cyclic.

how do u know its not completely true?
did u try drawing it?

 

[Xayma]

isnt there a initial condition of A,B,E,D being concyclic?

No they dont have to be.

Pfft sif use MS paint, CAD programs are great for diagrams for that.

 

[Archman]

oops, sorry, my fault, they don't have to be.

the question is fine.

 

[CrashOveride]

Yeah thats what i said, need to use vector-based program...paint doesn't cut it...thats why the diagram may show circles that appear non con-cyclic

Mojako is a poet.

 

[Archman]

no.. i proved it.

 

[mojako]

no.. i proved it.

Hereby I request thee to reveal the proof to me.

Mojako is a poet.

I am Roberd Frost.
Two roads diverged in a yellow wood.
O'
which one should I take??
I took the one less travelled by.
And I felt nothing different.

 

[Archman]

well other people may want more time to work on it.
'ere we go
let circle FBD and FAE intersect at C
ACB=BCF-ACF=BDF-FEA=DHE=AHB (they are angles)
so circle ABH goes through C
repeat argument interchanging D with A, E with B, and you've proved circle DEH goes through C. So all 4 goes through C.
oh forgot to say
warning: invisi text contains spoilers!(bah too late)

 

[mojako]

Thanks Archman!! ^_^
I can't see how other people may want more time to work on it though.
And if they want they can just not read the solution.
Thanks again.

 

[mojako]

Let circles ABH, HDE meet at X.

Then angle BXD = angle BXH + angle XHD = angle FAH + angle HED (exterior angle of cyclic ABXH; angles at circumference of HXED on arc DH) = 180 degrees - angle BFD (angle sum of AFE). So BXDF is cyclic (opposite angles are supplementary), so X lies on circle FBD.

Similarly X lies on circle FAE (the above argument, with A swapped with D and B swapped with E).

Thus X is a common point for the four circles.

see Archman,
this is becoz of you!!
wasting ppl's time.
but laurie did this in a split second anyway, so he's not disadvantaged.
Thanks laurie

BTW, "angle BXH + angle XHD" should be angle "BXH + angle HXD"

 

[ngai]

lol laurie's here now
hehe as i can see, hes not as experienced on these boards...
rite? ; )

 

[Xayma]

Why would you decide to start posting with <100 hours till the HSC
Now if you make continual remarks he will spend the 100 hours trying to figure out the colours

 

[nit]

maybe because you don't need to study anymore, and you're dead sure you'll get 100

 

[withoutaface]

maybe because you don't need to study anymore, and you're dead sure you'll get 100

Hahaha as I'm sure is the case for at least 3 people on this board

 

[Archman]

gday laurie. shouldn't add milk to tea all the time.

 

[mojako]

huh?

i dont get it.

IMO (in my opinion, or otherwise) participants of this forum should get it easily