Problem: Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter $AB$.
Denote by $P$, $Q$, $R$, $S$ the feet of the perpendiculars
from $Y$ onto lines $AX$, $BX$, $AZ$, $BZ$, respectively.
Prove that the acute angle formed by lines $PQ$ and $RS$ is half the size of $\angle XOZ$,
where $O$ is the midpoint of segment $AB$.

Solution: We present two possible approaches.
The first approach is just ``bare-hands'' angle chasing.
The second approach requires more insight but makes it clearer what is going on;
it shows the intersection point of lines $PQ$ and $RS$ is the foot
from the altitude from $Y$ to $AB$ using Simson lines.
The second approach also has the advantage that it works even if $\ol{AB}$
is not a diameter of the circle.

\paragraph{First approach using angle chasing.}
Define $T = \ol{PQ} \cap \ol{RS}$.
Also, let $2\alpha$, $2\beta$, $2\gamma$, $2\delta$ denote the measures of arcs
$\arc{AX}$, $\arc{XY}$, $\arc{YZ}$, $\arc{ZB}$, respectively,
so that $\alpha + \beta + \gamma + \delta = 90\dg$.

\begin{center}
\begin{asy}
pair A = dir(180);
pair B = dir(0);
pair X = dir(140);
pair Y = dir(100);
pair Z = dir(60);

pair P = foot(Y, A, X);
pair Q = foot(Y, B, X);
pair R = foot(Y, A, Z);
pair S = foot(Y, B, Z);

pair T = extension(P, Q, R, S);

draw(A--X--Y--Z--B, orange);
filldraw(unitcircle, opacity(0.1)+orange, red);

draw(Y--P, red);
draw(Y--Q, red);
draw(Y--R, red);
draw(Y--S, red);
draw(P--X--B, brown);
draw(P--T--S, red);
draw(S--Z--A, brown);

pair O = origin;
draw(arc(O, B, Z), blue+1.3, Margin(2,2));
draw(arc(O, Z, Y), blue+1.3, Margin(2,2));
draw(arc(O, Y, X), blue+1.3, Margin(2,2));
draw(arc(O, X, A), blue+1.3, Margin(2,2));

label("$2\alpha$", dir(160), dir(160), blue);
label("$2\beta$",  dir(122), dir(122), blue);
label("$2\gamma$", dir( 82), dir( 82), blue);
label("$2\delta$", dir( 30), dir( 30), blue);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$Z$", Z, dir(Z));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(210));
dot("$R$", R, dir(340));
dot("$S$", S, dir(S));
dot("$T$", T, dir(-90));

/* TSQ Source:

A = dir 180
B = dir 0
X = dir 140
Y = dir 100
Z = dir 60

P = foot Y A X
Q = foot Y B X
R = foot Y A Z
S = foot Y B Z

T = extension P Q R S

A--X--Y--Z--B orange
unitcircle 0.1 orange / red

Y--P red
Y--Q red
Y--R red
Y--S red
P--X--B brown
P--T--S red
S--Z--A brown

*/
\end{asy}
\end{center}
We now compute the following angles:
\begin{align*}
  \angle SRY &= \angle SZY = 90\dg - \angle YZA = 90\dg - (\alpha + \beta) \\
  \angle YQP &= \angle YXP = 90\dg - \angle BXY = 90\dg - (\gamma + \delta) \\
  \angle QYR &= 180\dg - \angle (\ol{ZR}, \ol{QX})
    = 180\dg - \frac{2\beta + 2\gamma + 180\dg}{2} = 90\dg - (\beta + \gamma).
\end{align*}
Hence, we can then compute
\begin{align*}
  \angle RTQ &= 360\dg - \left(
    \angle QYR + (180\dg - \angle SRY) + (180\dg - \angle YQP) \right) \\
  &= \angle SRY + \angle YQP - \angle QYR \\
  &= \left( 90\dg - (\alpha + \beta) \right) + \left( 90\dg - (\gamma + \delta) \right)
    - \left( 90\dg - (\beta + \gamma) \right) \\
  &= 90\dg - (\alpha + \delta) \\
  &= \beta + \gamma.
\end{align*}
Since $\angle XOZ = \frac{2\beta+2\gamma}{2} = \beta+\gamma$, the proof is complete.


\paragraph{Second approach using Simson lines, ignoring the diameter condition.}
In this solution, we will ignore the condition that $\ol{AB}$ is a diameter;
the solution works equally well without it,
as long as $O$ is redefined as the center of $(AXYZB)$ instead.
We will again show the angle formed by lines $PQ$ and $RS$ is half the measure of $\arc{XZ}$.

Unlike the previous solution, we instead define $T$ to be the foot from $Y$ to $\ol{AB}$.
Then the Simson line of $Y$ with respect to $\triangle XAB$ passes through $P$, $Q$, $T$.
Similarly, the Simson line of $Y$ with respect to $\triangle ZAB$ passes through $R$, $S$, $T$.
Therefore, point $T$ coincides with $\ol{PQ} \cap \ol{RS}$.

\begin{center}
\begin{asy}
pair A = dir(205);
pair B = -conj(A);
pair X = dir(140);
pair Y = dir(100);
pair Z = dir(60);

pair P = foot(Y, A, X);
pair Q = foot(Y, B, X);
pair R = foot(Y, A, Z);
pair S = foot(Y, B, Z);

pair T = extension(P, Q, R, S);

draw(A--X--Y--Z--B, orange);
filldraw(unitcircle, opacity(0.1)+orange, red);

draw(Y--P, red);
draw(Y--Q, red);
draw(Y--R, red);
draw(Y--S, red);
draw(P--X--B, brown);
draw(P--T--S, red);
draw(S--Z--A, brown);

draw(A--B, deepcyan+1.4);
draw(Y--T, deepcyan+1.4);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$Z$", Z, dir(Z));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(210));
dot("$R$", R, dir(340));
dot("$S$", S, dir(S));
dot("$T$", T, dir(-90));

/* TSQ Source:

A = dir 180
B = dir 0
X = dir 140
Y = dir 100
Z = dir 60

P = foot Y A X
Q = foot Y B X
R = foot Y A Z
S = foot Y B Z

T = extension P Q R S

A--X--Y--Z--B orange
unitcircle 0.1 orange / red

Y--P red
Y--Q red
Y--R red
Y--S red
P--X--B brown
P--T--S red
S--Z--A brown

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

Now it's straightforward to see $APYRT$ is cyclic (in the circle with diameter $\ol{AY}$),
and therefore
\[ \angle RTY = \angle RAY = \angle ZAY. \]
Similarly,
\[ \angle YTQ = \angle YBQ = \angle YBX. \]
Summing these gives $\angle RTQ$ is equal to half the measure of arc $\arc{XZ}$ as needed.