Circle Geometry Question (1 Viewer)

[AreYouAlright?]

Sorry for lack of diagram, my printer blew up when we got hit by lightning perhaps someone can find a solution to this question?

Two intersecting circles meet at F and C. The points B,C and D are collinear, the three chords produced from these points; BG, CF and DE are concurrent at A. Show that the quadrilateral AGFE is a cyclic quadrilateral.

I have no idea how to attempt this question, can anyone help?

Here is a rough drawing of the problem in paint:

 

[Jago]

where's points a, b, c, d?

try drawing it in paint or something.

 

[AreYouAlright?]

Added Drawing

Thanks for the advice, did a rough sketch in paint.. do you have any idea how to get the solution?

 

[Jago]

here's the solution, do you need any explanation?

 

[AreYouAlright?]

Explanation Plz...

Are you using the cyclic quadrilateral external angle property, and how does that prove that those points above are a cyclic quadrilateral?

Wait nevermind, lol i just figured out how you did it, yeah that's using cyclic quad ext angle. Thanks, i can never see which theorem to use when i look at those questions.. lol i should probably not to 4u next year.

 

[Jago]

haha i can never see it either, with this one i looked at it for 2 minutes straight then decided to pick out a theorem (external angle of triangle) and tried to prove it 1st time lucky

 

[AreYouAlright?]

Haha

LOL they aren't triangles tho..... u mean cyclic quadrilaterals don't u?

Also, while we are at it:

I don't have the soln to this previous exam and i don't know if i've answered this right so you might be able to help me...

3
------- < or = x
x-2

I got:

1 <or= x <or= -3 AND -3 <or= x <or= 1 but x can not = 2

should there be two cases? or is it meant to be just one and am i anywhere near right?

 

[Jago]

yep that's the one.

 

[AreYouAlright?]

Extra Question

I added an extra question there.. maybe u can help me with that one tooo

 

[Jago]

i get

-1< or = x <2

and

x > or = 3

 

[AreYouAlright?]

How'd you get that?

Do you solve that by squaring both sides... or do you do two cases?

 

[Jago]

squaring both sides.

 

[AreYouAlright?]

Ahh ok.. i know what i did wrong now

It's ok i just forgot to invert the signs i.e (x-3)(x+1)=0
x= +3 or -1

i was using -3 or 1
so that was giving me the wrong answer but it's k now... thanks for your help

 

[jumb]

squaring both sides.

I thought you timesed both sides by the square of the denominator?

and then simplify it.

 

[Jago]

that's what i meant