[asy][asy] import olympiad; size(8cm); defaultpen(fontsize(10pt)); pair A=dir(-29), B=dir(209), C=dir(72); pair A_1 = dir(310), B_1=dir(-29-137), C_1=dir(108); pair A_2 = dir(-29-180), B_2=dir(209-180), C_2=dir(72-180); draw(circumcircle(A,B,C)); draw(A--B--C--A); draw(A--A_1--A_2, blue+dashed); draw(B--B_1--B_2, blue+dashed); draw(C--C_1--C_2, blue+dashed); label("$A$",A,(1,-1)); label("$B$",B,(-1,-1)); label("$C$",C,(1,1)); label("$A_1$",A_1,(1,-1)); label("$B_1$",B_1,(-1,-1)); label("$C_1$",C_1,(0,1)); [/asy][/asy] Firstly, note that since $\overline{AA_1} \parallel \overline{BC}$ and $A_1 \in (ABC)$, we get that $AA_1BC$ is an isosceles trapezoid. Similarly, $ABB_1C$ and $ABC_1C$ are also isosceles trapezoids. Now, by Carnot's theorem, the perpendiculars from $A_1$, $B_1$, $C_1$ to $\overline{BC}$, $\overline{AC}$ and $\overline{AB}$ respectively are concurrent if and only if $${AC_1}^2-{BC_1}^2+{A_1B}^2-{A_1C}^2+{B_1C}^2-{B_1A}^2 = 0$$. Since $AA_1BC$ is an isosceles trapezoid, we know that its legs are the same length and its diagonals are the same length. This means that $A_1B = AC$ and $A_1C = AB$. Similarly, for the other two isosceles trapezoids, we get that $B_1C = AB$, $AB_1 = BC$, $BC_1 = AC$ and $BC = AC_1$. Substituting these back into the expression, we get that $$ {AC_1}^2-{BC_1}^2+{A_1B}^2-{A_1C}^2+{B_1C}^2-{B_1A}^2 = {BC}^2-{AC}^2+{AC}^2-{AB}^2+{AB}^2-{BC}^2 = 0$$ which means that the perpendiculars from $A_1$, $B_1$, $C_1$ to $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ respectively are concurrent as desired.

Can't you just say that this concurrency point is the isotomic conjuate of the orthocenter? I don't know where I got the idea from but I'm pretty sure i got it from @above