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I have problems in solving the following problem:

Consider two circles which have only one point $A$ in common, i.e. which are tangent to each other. Now consider two lines through A, such that the lines meet the circles at further points $B,C,D,E$. I want to prove that the lines $DE$ and $BC$ are parallel lines ($D,E$ being points on one circle and $B,C$ on the other).

I tried to use theorems like the inscribed angle theorem, but I was not succesfull so far. Does someone know how to solve this problem?

If it is possible, I only want to use geometric arguments, and not analytical arguments. Does this result have any name?

Best wishes

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  • $\begingroup$ This theorem is true if the circles are of the same size. $\endgroup$ – Jack Jun 2 '15 at 6:51
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Hint:

Let $M,N$ two points on the common tangent at opposite sides with respect to $A$, than $\angle DAM = \angle DEA$ because they subtend the same arc $DA$ ( $\angle DAM$ is a ''limit'' angle, being the side $AM$ tangent, but, if you don't like the concept of ''limit'' angle, you can proof the same claim as a consequence of the tangent-secant theorem.).

In the same manner $\angle BAN=\angle BCA$.

Now note that $\angle DAM$ and $\angle BAN$ are opposed, and .....

I don't know if this result has a special name.

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Problem Figure.

Let the common tangent $L$ be drawn. Then, by the Alternate Segment Theorem: $$\angle DEA =\angle (DA,L)$$ And similarly: $$\angle CBA=\angle (CA,L)$$ But the latter are equal as they are vertically opposite. Therefore it follows that the angles Intercepted by $BE$ are equal and thus, the lines are parallel.


This is a special limiting case of the Reim's Theorem, which states that lines through the points of intersection of two circles cut the circle in points that form two parallel segments.

This is not a well-known theorem by the way.

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