Problem: Quadrilateral $XABY$ is inscribed in the semicircle $\omega$ with
diameter $\ol{XY}$.
Segments $AY$ and $BX$ meet at $P$.
Point $Z$ is the foot of the perpendicular from $P$ to line $\ol{XY}$.
Point $C$ lies on $\omega$ such that line $XC$ is perpendicular to line $AZ$.
Let $Q$ be the intersection of segments $AY$ and $XC$.
Prove that
\[\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.\]

Solution: Let $\beta = \angle YXP$ and $\alpha = \angle PYX$ and set $XY = 1$.
We do not direct angles in the following solution.

\begin{center}
\begin{asy}
size(10cm);
pair X = Drawing("X", dir(180), dir(225));
pair Y = Drawing("Y", dir(0), dir(315));
pair A = Drawing("A", dir(130), dir(135));
pair B = Drawing("B", dir(94), dir(94));
pair P = Drawing("P", IP(X--B,A--Y), dir(90));
pair Z = Drawing("Z", foot(P,X,Y), dir(-90));
pair C = 2 * foot(origin, A+Y-Z, Y) - Y;
Drawing("C", C, C);
pair Q = Drawing("Q", IP(A--Y,X--C), dir(80));
draw(X--A--B--Y--cycle);
draw(A--Y--C--X--B);
draw(P--Z--A);
draw(arc(origin,1,0,180));
markangle("$\beta$", Y, X, P);
markangle("$\alpha$", A, Y, X);
\end{asy}
\end{center}

Observe that
\[ \angle AZX = \angle APX = \alpha + \beta \]
since $APZX$ is cyclic.
In particular, $\angle CXY = 90^\circ- (\alpha + \beta)$.
It is immediate that
\[ BY = \sin \beta, \quad CY = \cos \left( \alpha + \beta \right),
    \quad AY = \cos \alpha, \quad AX = \sin \alpha.  \]
The Law of Sines on $\triangle XPY$ gives $XP = XY \frac{\sin\alpha}{\sin(\alpha+\beta)}$,
and on $\triangle XQY$ gives $XQ = XY \frac{\sin\alpha}{\sin(90+\beta)} = \frac{\sin\alpha}{\cos\beta}$.
So, the given is equivalent to
\[ \frac{\sin\beta}{\frac{\sin\alpha}{\sin(\alpha+\beta)}}
  + \frac{\cos(\alpha+\beta)}{\frac{\sin\alpha}{\cos\beta}}
  = \frac{\cos\alpha}{\sin\alpha} \]
which is equivalent to $\cos\alpha = \cos\beta\cos(\alpha+\beta) + \sin\beta\sin(\alpha+\beta)$.
This is obvious, because the right-hand side is just $\cos\left( (\alpha+\beta)-\beta \right)$.