Here's a pretty quick proof of Droz-Farnzy Theorem:
Droz-Farnzy Theorem wrote:
Let ABC be a triangle with orthocenter H. Let l_1,l_2 be any two perpendicular lines passing through H. Let (l_1,l_2) intersect lines BC,CA,AB at points (X,X'),(Y,Y'),(Z,Z'), respectively. Then midpoints of segments XX',YY',ZZ' are collinear.
When \triangle ABC is right-angled, the problem is trivial (assume otherwise henceforth). We want centers of \odot(HXX'),\odot(HYY'),\odot(HZZ') to be collinear, which is equivalent to the three circles being coaxial. Define M as the Miquel point of lines BC,CA,AB,l_1. Clearly M \ne H since M \in \odot(ABC) while H \notin \odot(ABC). Now the reflection of M in lines BC,CA,AB,l_1 are collinear on some line \tau. Then H \in \tau as M \in \odot(ABC). Since l_1 \perp l_2, so reflection of any point in lines l_1,l_2 are collinear with H. It basically follows that reflection of M in the five lines BC,CA,AB,l_1,l_2 (and point H) are collinear. So M is the Miquel point of the five lines, which means all of \odot(HXX'),\odot(HYY'),\odot(HZZ') pass through M. So the three circles intersect at another point distinct from H, hence they are coaxial. \blacksquare