Problem: Let $ABCD$ be a cyclic quadrilateral.
Let $P, Q$ and $R$ be the feet of perpendiculars
from $D$ to lines $\ol{BC}$, $\ol{CA}$ and $\ol{AB}$, respectively.
Show that $PQ = QR$ if and only if the
bisectors of angles $ABC$ and $ADC$ meet on segment $\ol{AC}$.

Solution: Let $\gamma$ denote the circumcircle of $ABCD$.
The condition on bisectors is equivalent to $(AC;BD)_\gamma = -1$.
Meanwhile if $\infty$ denotes the point at infinity along Simson line $\ol{PQR}$
then $PQ = QR$ if and only if $(PR;Q\infty) = -1$.

Let rays $BQ$ and $DQ$ meet the circumcircle again at $F$ and $E$.

\begin{center}
\begin{asy}
pair B = dir(150);
pair A = dir(200);
pair C = dir(340);
pair D = IP(2*A*C/(A+C)--B, unitcircle);

draw(A--D--C--B--A--C, lightblue);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
pair P = foot(D, B, C);
pair Q = foot(D, A, C);
pair R = foot(D, A, B);
draw(P--D--R, blue);
draw(R--A, lightblue);

draw(P--R, red);
pair O = origin;
pair E = -D+2*foot(O, D, Q);
pair F = -B+2*foot(O, B, Q);

draw(B--E, red);
draw(B--F, lightblue);
draw(D--E, lightblue);

dot("$B$", B, dir(B));
dot("$A$", A, dir(A));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$P$", P, dir(60));
dot("$Q$", Q, dir(315));
dot("$R$", R, dir(R));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));

/* TSQ Source:

B = dir 150
A = dir 200
C = dir 340
D = IP 2*A*C/(A+C)--B unitcircle

A--D--C--B--A--C lightblue
unitcircle 0.1 lightcyan / blue
P = foot D B C R60
Q = foot D A C R315
R = foot D A B
P--D--R blue
R--A lightblue

P--R red
O := origin
E = -D+2*foot O D Q
F = -B+2*foot O B Q

B--E red
B--F lightblue
D--E lightblue

*/
\end{asy}
\end{center}

\begin{lemma*}
  [EGMO Proposition 4.1]
  Then $\ol{BE} \parallel \ol{PQR}$.
\end{lemma*}
\begin{proof}
  Since $\dang DPR = \dang DAR = \dang DAB = \dang DEB$.
\end{proof}

Now we have
\[ (PR;Q\infty) \overset{B}{=} (CA;FE)_\gamma
  \overset{Q}{=} (AC;BD)_\gamma \]
as desired.