USAMO 2005, Problem 3: Let $ABC$ be an acute-angled triangle, and let $P$ and $Q$ be two points on its side $BC$. Construct a point $C_{1}$ in such a way that the convex quadrilateral $APBC_{1}$ is cyclic, $QC_{1}\parallel CA$, 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, $QB_{1}\parallel BA$, 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.
Here is a solution I found for the problem mentioned above:
Define $X=C_1Q \cap AB$ and $Y=B_1Q \cap AC$. Since $AX \parallel QY$ and $AY \parallel QX, AXQY$ is a parallelogram $(1)$.
By power of point $X$ in $(APB)$ and of point $Y$ in $(APC)$ we have $XA\times XB=XC_1 \times XQ$ and $YA \times YC= YB_1 \times YQ$. Therefore $\frac{XB}{XC_1}=\frac{XQ}{XA}=\frac{AY}{QY}=\frac{YB_1}{YC}$ (using $AX=QY$ and $XQ=AY$ derived from $(1)$ ). $(2)$ Also, from $(1)$ we have that $\angle{C_1XB}=\angle{CYB_1}$ and by combining this with $(2)$ we have that $\triangle{C_1XB}$ and $\triangle{CYB_1}$ are similar. Since the two pairs of corresponding sides in this similar triangles are parallel, the third pair is parallel too. i.e., $C_1B\parallel B_1C$ $(3)$. Note that $\angle{BC_1A}=\angle{APC}=180^{\circ}- \angle{AB_1C}$ and from $(3)$ $C,A,B$ collinear.
Then the result follows, since $\angle B_1C_1P=\angle AC_1P=\angle ABP=\angle B_1QC=\pi-\angle B_1QP$.
Am I doing something wrong? (this solution seems pretty easy, and nobody noticed it before. That makes me question its validy)