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
Best Answer
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.