Problem: Let $ABC$ be a triangle and $D$, $E$, $F$ be the midpoints of arcs $BC$, $CA$, $AB$ on the circumcircle.
Line $\ell_a$ passes through the feet of the perpendiculars
from $A$ to $\ol{DB}$ and $\ol{DC}$.
Line $m_a$ passes through the feet of the perpendiculars from $D$ to $\ol{AB}$ and $\ol{AC}$.
Let $A_1$ denote the intersection of lines $\ell_a$ and $m_a$.
Define points $B_1$ and $C_1$ similarly.
Prove that triangles $DEF$ and $A_1B_1C_1$ are similar to each other.

Solution: In fact, it is true for any points $D$, $E$, $F$ on the circumcircle.
More strongly we contend:
\begin{claim*}
  Point $A_1$ is the midpoint of $\ol{HD}$.
\end{claim*}
\begin{proof}
  Lines $m_a$ and $\ell_a$ are Simson lines,
  so they both pass through the point $(a+b+c+d)/2$
  in complex coordinates.
\end{proof}

\begin{center}
  \begin{asy}
    pair A = dir(130);
    pair B = dir(200);
    pair C = dir(-20);
    pair D = dir(-90);
    pair P1 = foot(A,B,D);
    pair P2 = foot(A,C,D);
    pair Q1 = foot(D,A,B);
    pair Q2 = foot(D,A,C);
    pair H = A+B+C;
    draw(H--D, deepgreen);
    draw(unitcircle);
    filldraw(A--B--C--cycle, lightcyan + opacity(0.1), blue);
    draw(B--D--C, blue);
    draw(P1--A--P2, dashed+red);
    draw(Q1--D--Q2, dashed+orange);
    draw(P1--P2, red);
    draw(Q1--Q2, orange);
    draw(P1--B, blue);
    draw(Q1--B, blue);
    dot(Q1, orange);
    dot(Q2, orange);
    dot(P1, red);
    dot(P2, red);
    dot("$A_1$", (A+B+C+D)/2, dir(-90));
    dot("$A$", A, A);
    dot("$B$", B, B);
    dot("$C$", C, C);
    dot("$D$", D, D);
    dot("$H$", H, dir(90));
  \end{asy}
\end{center}

Hence $A_1 B_1 C_1$ is similar to $DEF$ through a homothety at $H$ with
ratio $\half$.