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. $\blacksquare$