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**Circles #3**

1. If from any point on the extended common chord of 2 intersecting circles, tangents be drawn to the circles, prove that they are equal.

*Last edited by Agnishom (2014-01-30 00:17:32)*

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,926

Just read this. Give me a moment and I'll see if I can do this.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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Just read what?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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Hi bob;

I explored the problem with GeoGebra and I fohund a hint. I think it will solve the problem, but I do not understand how to put it to words. Will you see my figure?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,926

I was working on sundarrd's post so I hadn't spotted yours until that moment ... hence just read it.

I've got a diagram and checked out the measurements. I've left in my construction lines (in red) as I think it may help.

AB is perpendicular to CD.

Still thinking .....................

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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Hint: Through EF and FG draw perpendiculars meeting at K, now draw a circle with K as center and EF and GF tangents.

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,926

I've put K at the intersection of FA and GB. But there's still a bit to prove. Are you saying that you've done this now?

Bob

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No, I still do not understand how to tell that KF and KG are equal.

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,926

I did this:

Bisect angle FEG.

Extend FA to cut this bisector at K

Draw a circle, centre K and radius KF.

It looks like KBG is a straight line. If it is, then the equal tangent rule applies and we're done.

But why should KBG be straight?

Bob

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Well, can you prove this: For every pair of distinct intersecting lines, there exists at least one circle of which these lines are tangents.

If you can prove that, we can do the following:

Assume that the circle of which EF and FG are tangents has already been drawn. Now, Since they are the tangents drawn from the same external point on the circle, they are equal.

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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bob bundy wrote:

But why should KBG be straight?

Because it is supposed to be an extended radii of the jumbo circle.

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,926

Hhmmm. Not sure this helps.

see also diagram below.

Draw any lines FE and GE.

Bisect angle FEG.

Choose K anywhere on this bisector.

Construct MK perpendicular to FE with M on FE.

Draw circle, centre K and radius KM.

Mark N on GE so that EM = EN.

It is fairly easy to show using congruent triangles that KN is perpendicular to GE.

So I have drawn a circle as you require.

In fact, there are an infinite number of possible circles because K can be anywhere on the bisector (except at E).

So slide K along the bisector until KN coincides with KG.

But M is not at F.

Bob

Got to go and do some woodwork now. Back later.

*Last edited by bob bundy (2014-01-30 02:02:09)*

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Will you chop trees?

Sorry, on pen and paper we can't slide K.

Can you prove this statement? *Given a pair of distinct intersecting lines there exists at least one circle of which they are tangents.*

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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Another proof:

Draw the tangent EH on the circle which contains point F. Now, Draw a circle whose tangents from E are EG and EH.

Now, EG=EH; EH=EF ⇒ EG=EF [QED]

The big question is whether *a circle whose tangents from E are EG and EH* is possible.

*Last edited by Agnishom (2014-01-30 04:14:36)*

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,926

hi,

No tree chopping! This was more delicate. Fixing some drawers where the bottom was coming off and then making a device for washing windows.

It must have been worth doing because I think I now have a proof.

Construct K as follows:

Make the angle bisector for angle E and extend FA to intersect this line at K.

Consider the triangles KFE and FGE.

They have o ... drat that won't work.

Back to the drawing board.

Bob

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Hi;

Did you look at post 14? It looks more convincing even though it does not tell us anything on how that circle could be drawn

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,926

I shall take a little break to let my subconscious get to work, and then come back to try that. It will be diagram number 6 so far on this problem.

Don't go and do it while I'm gone.

Bob

*Last edited by bob bundy (2014-01-30 04:58:08)*

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,053

Hi,

I was getting nowhere and so went surfing instead, whereupon I found this theorem on the net (I changed the point letters to match the ones in posts above):

Theorem:

If E is a point outside a circle and E,D,G are points on the circle such that EG is a tangent and EDC is a secant, then EG² = ED x EC

Proof:

∠EGD = ∠GCD (alternate segment theorem)

∠EGC = ∠GDE (angle sum of a triangle)

∴ ΔEGC is similar to ΔEDG

∴ EG/ED = EC/EG, from which EG² = ED x EC

Similarly, EF² = ED x EC, and ∴ EF = EG

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,926

Oh yes! And I knew that theorem too. Ggrr. If only I had gone surfing. But the sea temperature is around 9 C. Ugh!

Bob

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Alternate segment theorem?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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Don't go and do it while I'm gone.

Sorry, I'll have to sleep.

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,129

Can you prove this statement? Given a pair of distinct intersecting lines there exists at least one circle of which they are tangents.

Let's try by constructing a general example of one.

1) Draw two lines that intersect. We can obviously always do this. Call the point of Intersection C.

2) Choose the acute angle or one of the right angles and draw the angle bisector. We can also always do this.

3) Mark off point B on the first line and point A on the second line so that angle ACB is 90 degrees or less. In other words choose to place them in such a way that we are dealing with the acute or right angle. Also make the distance from C to A equal the distance from C to B. We can always do this.

4) Draw perpendicular lines through B and C and call the point where they intersect the angle bisector O.

5) Triangles COB and COA are congruent because of SAS. Therefore OA = OB.

6) O is the center of the circle that passes through A and B and therefore the two original intersecting lines are tangent to it. Since these steps can always be accomplished there is always one circle tangent to any two intersecting lines.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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Please look at post 1 and 14. Do they look ok?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,129

Hi Agnishom;

I do not know.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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4) Draw perpendicular lines through B and C and call the point where the intersect the angle bisector O.

Hwc can you tell for sure if those lines have the same meeting point?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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