Problem: Let $ABC$ be an acute-angled triangle,
and let $P$ and $Q$ be two points on side $BC$.
Construct a point $C_1$ in such a way that the convex quadrilateral $APBC_1$ is cyclic,
$\ol{QC_1} \parallel \ol{CA}$, and $C_1$ and $Q$ lie on opposite sides of line $AB$.
Construct a point $B_1$ in such a way that the convex quadrilateral $APCB_1$ is cyclic,
$\ol{QB_1} \parallel \ol{BA}$, and $B_1$ and $Q$ lie on opposite sides of line $AC$.
Prove that the points $B_1$, $C_1$, $P$, and $Q$ lie on a circle.

Solution: It is enough to prove that $A$, $B_1$, and $C_1$ are collinear,
since then $\dang C_1QP = \dang ACP = \dang AB_1P = \dang C_1B_1P$.

\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair P = 0.7*C+0.3*B;
draw(A--B--C--cycle);

pair O_1 = circumcenter(A, P, C);
pair B_1 = 2*O_1-C;
pair Q = extension(B, C, B_1, B_1+B-A);
pair C_1 = extension(A, B_1, Q, Q+C-A);

draw(circumcircle(A, B, P));
draw(circumcircle(A, C, P));

draw(C_1--Q--B_1);
draw(B_1--C_1, dotted);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$P$", P, dir(P));
dot("$B_1$", B_1, dir(B_1));
dot("$Q$", Q, dir(Q));
dot("$C_1$", C_1, dir(C_1));

/* Source generated by TSQ */
\end{asy}
\end{center}

\paragraph{First solution.}
Let $T$ be the second intersection of $\ol{AC_1}$ with $(APC)$.
Then readily $\triangle PC_1T \sim \triangle ABC$.
Consequently, $\ol{QC_1} \parallel \ol{AC}$ implies $TC_1QP$ cyclic.
Finally, $\ol{TQ} \parallel \ol{AB}$ now follows from the cyclic condition,
so $T = B_1$ as desired.

\paragraph{Second solution.}
One may also use barycentric coordinates.
Let $P = (0,m,n)$ and $Q = (0,r,s)$ with $m+n=r+s=1$.
Once again, \[ (APB) : -a^2yz-b^2zx-c^2xy + (x+y+z)(a^2m \cdot z) = 0.  \]
Set $C_1 = (s-z, r, z)$, where $C_1Q \parallel AC$ follows by $(s-z) + r + z = 1$.
We solve for this $z$.
\begin{align*}
  0 &= -a^2rz + (s-z)(-b^2z-c^2r) + a^2mz \\
  &= b^2z^2 + (-sb^2+rc^2)z - a^2rz + a^2mz - c^2rs \\
  &= b^2z^2 + (-sb^2+rc^2+a^2(m-r))z - c^2rs \\
  \implies 0 &= rb^2 \left( \frac{z}{r} \right)^2 + (-sb^2+rc^2 + a^2(m-r)) \left(\frac{z}{r}\right) - c^2s.
\end{align*}
So the quotient of the $z$ and $y$ coordinates of $C_1$ satisfies this quadratic.
Similarly, if $B_1 = (r-y, y, s)$ we obtain that
\[
0 = sc^2 \left( \frac{y}{s} \right)^2 + (-rc^2 + sb^2 + a^2(n-s)) \left(\frac{y}{s}\right) - b^2r
\]
Since these two quadratics are the same when one is written backwards (and negated),
it follows that their roots are reciprocals.
But the roots of the quadratics represent $\tfrac{z}{y}$ and $\tfrac{y}{z}$
for the points $C_1$ and $B_1$, respectively.
This implies that the points $B_1$ and $C_1$ are collinear with $A=(1,0,0)$
(in some line of the form $\tfrac{y}{z} = k$), as desired.