Seven Circles Theorem Proof
Six circles , , , , , , lie tangent around a seventh, . Each is also tangent with the next in sequence, and the intersection points they form with , , , , , , and , lie along in that order. In that case, , , and are concurrent.
Lemma 1: , the tangent to at , and are all concurrent.
Let the second touch points of at and be and , respectively. By angle chasing, is cyclic:
Similarly we get that . Therefore
By radical axis theorem on , , and , , , and the tangent to at all intersect at the same point. Call this point , and define and similarly.
Let and be the second intersection of and with , respectively.
By angle chasing it is easy to show that . This means that is similar to by homothety at . Next, . This also means since and are both isosceles, and so is similar to . Therefore the homothety at that takes to also takes to , and it follows is collinear with .
This allows us to redefine as the intersection of with . Now, by Desargues' theorem, we have two triangles in perspective from a point: , and . Therefore we will have that are collinear.
By Menalaus' theorem on and line , we will have . We multiply this by to get . But this is equal to , so we have , which means .
Now I claim this condition implies that , , and all intersect. Let be the intersection of and , and let be the second intersection of with . Then, we have . This implies that . Now there are two possibilities for , but only one of them will lie on the "right" side of segment (i.e. the segment which does not contain ,, and ). Below is a diagram of on the wrong side, and you can see that the diagonals don't line up. If we restrict 's placement to the right side of , then it is fairly clear there is only one point on such that : if there were 2 valid points and , then we would WLOG have < , and > simultaneously, which would imply < , a contradiction.