Given a scalene triangle $\triangle ABC$ and $D,E,F$ the midpoints of $BC, CA, AB$ and $A_1, B_1, C_1$ points such that $\angle AA_1C=90^{\circ}$ and $\angle BB_1C=90^{\circ}$ and $\angle CC_1A=90^{\circ}$. $A_2$ is the intersection of $BC$ and $B_1C_1$. We define $B_2$ and $C_2$ in the same manner. Prove that the lines which from $D,E,F$ which form a right angle with $AA_2, BB_2, CC_2$ respectively, coincide.
We can easily prove that they all pass through the orthocenter $H$ of the triangle, withe use of the Brocard theorem My question is, is there any other way to prove this?
Brokard's theorem is the natural way, but here's a more elementary approach.
Let $(AB_1C_1)$ meet $(ABC)$ again at $G$. The radical axis theorem on $(AB_1C_1)$, $(ABC)$ and $(BCB_1C_1)$ implies that $G$ lies on $AA_2$.
It is known that the reflection $H'$ of $H$ over $D$ is diametrically opposite $A$ on $(ABC)$ (prove it!). So $\angle AGH=\angle AGH'=90^\circ$, which implies $G$, $H$, $D$, $H'$ are collinear.
Similarly, the other two lines also pass through $H$.