USAMO 2005, Problem3 (Triangle Geometry)- Is my solution correct? 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)
 A: Let $\ell$ be an arbitrary line through $A$ not passing through the interior of triangle $ABC$. Let $B_3$ and $C_3$ be the second intersections (other than $A$) of $\ell$ with the circumcircles of triangles $APC$ and $APB$, respectively; these points both lie on the same side of the line $BC$ as does $A$. (note that one of these intersections may coincide with $A$ in case $\ell$ is tangent to one of the two circles.) Let $Q_1$ be the second intersection (other than $P$) of the line $BC$ with the circumcircle of triangle $B_3C_3P$. (Again, this may coincide with $P$ in case $BC$ is tangent to the circle at $P$.) In case $Q_1$ lies between $B$ and $P$, we have$$\angle C_3Q_1P = 180^\circ - \angle PB_3C_3 = 180^\circ - \angle PB_3 A = \angle ACP,$$by noting that $B_3C_3Q_1P$ and $AB_3CP$ are cyclic; in case $P$ lies between $B$ and $Q_1$, we have$$\angle C_3 Q_1P = \angle C_3B_3P = \angle AB_3P = \angle ACP$$by noting that $B_3C_3PQ_1$ and $AB_3CP$ are cyclic. In either case, $C_3Q_1$ is parallel to $AC$; similarly, $B_3Q_1$ is parallel to $AB$.

In case $Q = Q_1$, the points $C_1$ and $C_3$ both lie on the line through $Q$ parallel to $AC$, and both lie on the same side of $BC$ as does $A$. There is only one such point, so $C_1 = C_3$; similarly $B_1 = B_3$. In particular, if $Q = Q_1$, then the points $B_1 = B_3$, $C_1 = C_3$, $P$, $Q = Q_1$ lie on a circle.
It thus suffices to show that there exists some $\ell$ such that $Q = Q_1$. To see this, note that $\ell$ may be rotated continuously between the position where it coincides with $AB$, in which case $C_3 = Q_1 = B$, and the position where it coincides with $AC$, in which case $B_3 = Q_1 = C$. As $\ell$ rotates, $Q_1$ moves continuously along the line $BC$, so at some point it must pass through $Q$. As noted above, this implies the desired result.
Remark. The point $Q_1$ need not move directly from $B$ to $C$; it may first move in the opposite direction (away from $C$) before changing course and heading towards $C$. Likewise, it may overshoot $C$ before finally settling at $C$. All that matters is that the motion is continuous, so the point $Q$ must be passed through somewhere along the way.
