Suppose A, B, C are arbitrary points on a straight line and X is a point not on the line. Construct similar and similarly oriented triangles ABX and BCY. On line BX choose a point X', on line BY choose a point Y'. If triangle X'Y'Z' is similar to triangles ABX and BCY but with a different orientation then Z' is always collinear with A, B, and C!
This beautiful result and its proof are due to Nathan Bowler. This is a clear generalization of the case where X' = X and Y' = Y.
Three Similar Triangles
Let the origin O coincide with B. Define a function F so that for any P and Q, F(P, Q) is the unique point R with PQR similar to ABX but with a different orientation. (So that, e.g., Z = F(X, Y).) F is clearly a linear function. Now define a function G of pairs of scalars (a, b) by G(a, b) = F(aX, bY). G is also clearly linear.
In particular, the image of G is either the whole plane or a straight line through O (it is not the point O since X and Y are nonzero). Now let k be such that B = F(X, kY). Note that there must be such a k since 
AXB = YBX. Then B = G(1, k) and, since also B = G(0, 0), G is not a bijection. Therefore the image of G is a straight line through B.
Now, let R = G(1, 0). Since R = F(X, B), RBX = ABX, so that R lies on AB. Thus the image of G is AB. This exactly means that, for any a and b, G(a, b) lies on that line, i.e. is collinear with A, B, and C.
In particular, the foregoing implies a specific case where X' = X and Y' = Y, such that Z' = Z = F(X, Y). Indeed, in this case G(1, 1) = Z so Z lies on AB.
Copyright © 1996-2008 Alexander Bogomolny
