Agnishom wrote:

What about bobbym's proof? He still did not tell me why those points must coincide.

I think that is the easiest part, they are constructed to coincide. See the other thread.

]]>Ggrr. If only I had gone surfing. But the sea temperature is around 9 C. Ugh!

The water temperature is very bearable down here!

But the air temp isn't! We've had quite a few 40+ days lately, so the aircond's been working flat out (and thank goodness it's working!!)

The beach pics are ones I took of Bondi (Sydney) last November.

]]>(It is still cheating, because we do not know if that can be done.)

What about bobbym's proof? He still did not tell me why those points must coincide.

]]>I repeated the construction but with E not on CB.

The diagram is somewhat complicated so I've tried to colour code it. The dotted circles are just there to fix the points F and G.

H is the second tangent point as before. The red circle does have both the line EH and the line EG as tangents but not both of G and H are the actual tangent contact points. I've chosen to make G one such point and you can see that H is not the other. Big red arrow.

This just shows how careful you have to be proving things. The result is true when E is on CB so that diagram looks ok. But this diagram shows that the construction is in-valid.

Bob

]]>Anyway here's what I did.

(i) Join A to E, construct the midpoint and hence construct a circle with AE as diameter which makes EF a tangent to the original circle.

(ii) Do the same with the second circle to get G.

(iii) Find H the other point on the first circle which will make a tangent EH.

(iv) Bisect angle HEG.

(v) Find J where this bisector cuts BG.

(vi) Construct the circle with J as centre and with EG and EH as tangents.

Then EF = EH and EG = EH => EF = EG as required.

Bob

]]>Alternate segment theorem?

Haven't used this for years.

EDIT: earlier error corrected.

In any circle, centre A, EF a tangent at F, D some point on the circle:

Let EFD = x => FCD = x (5th angle property of a circle)

http://www.mathisfunforum.com/viewtopic … 88#p220488

Now consider the triangles EFC and EDF

angle E is common, and EFD = FCE = x

So EFC and EDF are similar

=> ED/EF = EF/EC => ED.EC = EF^2

Similarly ED.EC = EG^2

=> EF = EG

Thanks Phro.

Bob

]]>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?

]]>I do not know.

]]>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.

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

Sorry, I'll have to sleep.

]]>Bob

]]>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

]]>