Circle k(S; r)is touching point A of line AB. Circle l(T; s)is touching point B of line AB and intersects circle k in the edge points C, D of its diameter. Prove that the intersection M of lines CD and AB is the centre of line AB.

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- Jan 8th 2010, 04:28 AMx-matherProving . . .
Circle k(S; r)is touching point A of line AB. Circle l(T; s)is touching point B of line AB and intersects circle k in the edge points C, D of its diameter. Prove that the intersection M of lines CD and AB is the centre of line AB.

- Jan 8th 2010, 01:29 PMHenryt999Trying to help but...
Hi!

Reading the problem over and over and canīt quite grash it. C;D are antipods of large circle? And inside that circle are two other circles? - Jan 9th 2010, 07:41 AMx-mather
See this (I know it is not accurate but it helps)

http://content.imagesocket.com/thumbs/sssssss723.JPG - Jan 10th 2010, 05:03 AMx-mather
If you know it, written form will be sufficient for me.

- Jan 10th 2010, 05:50 AMArchie Meade
Here is a sketch of the geometry.

The large blue circle has the same radius as circle k.

Using the radius of the smaller circle,

next draw right-angled triangles to show |TF|=|TG|.

The show |BM|=|BA|

I had to shrink my sketch and it has become a little skewed.

Those lines are meant to be perpendicular.

This only works if CD is a centreline. - Jan 10th 2010, 05:54 AMHenryt999Well, does this make sence
Is this somewhat understandable?

- Jan 10th 2010, 06:07 AMArchie Meade
Simplest is the following...

Triangle TBE is isosceles,

from this you can show |BM|=|MA|

Again, CD needs to be a centreline of the smaller circle. - Jan 10th 2010, 07:06 AMArchie Meade
A general proof, when CD is not a diameter of the smaller circle

can be derived from the attached sketch. - Jan 10th 2010, 07:16 AMArchie Meade
Here is a geometric proof.