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**nghoihin1****Guest**

Thanks for your help!

Two circles intersect at points A and B. A common tangent touches the 1st circle at point C and the second at point D. Let B be inside the triangle ACD. Let the line CB intersect the second circle again at point E. Prove that AD bisects the angle CAE.

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

A common tangent touches the 1st circle at point C and the second at point D.

Does this mean that the line is tangent to both circles? If that's the case, then I think there are only two possible places for such a line. Either tangent to the very top, or very bottom.

Edit: nevermind, I made the assumption that the circles were the same size.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

heres an image representing what was said

*Last edited by luca-deltodesco (2006-08-17 23:33:31)*

The Beginning Of All Things To End.

The End Of All Things To Come.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Right. Notice that everything is not regular, i.e. no right angles, no lines are bisected, nothing is parrallel, etc. This means we got just about nothing to work with.

The only thing we really have to work with are the radii of the circles. I believe that using each radius, we can figure out things about angles.

Hopefully I'll have time to work on it later.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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