Proof of cyclicity using power of a point

An acute-angled triangle $ABC$ is given in the plane. The circle with diameter $AB$ intersects
altitude $CC’$ and its extension at points $M$ and $N$, and the circle with diameter $AC$ intersects altitude $BB’$ and its extensions at $P$ and $Q$. Prove that the points $M, N, P, Q$ lie on a common circle. (USAMO, 1990)

$MPNQ$ is cyclic if and only if $CBC’B’$ is cyclic. This is because $MPNQ$ is cyclic if and only if $MXcdot XN=PXcdot XQ$ by the power of $X$ on the circle whose existence is to be proven. Notice that $B’$ is on the circle with diameter $AB$, because $angle AB’B=90^{circ}$, thus $B’Xcdot XB=MXcdot XN.$

Also, $PXcdot XQ=CXcdot XC’,$ because $C’$ must be on the circle with diameter $AC$. Therefore, $MXcdot NX=PXcdot QXLongleftrightarrow B’Xcdot XB=CXcdot XC’$, which happens if and only if $CBC’B’$ is cyclic.

$CBC’B’$ is cyclic. This is because $angle B’CC’=angle C’BB’$, since $triangle ACC’simtriangle ABB’.$ Therefore, $MPNQ$ is also cyclic.

I think my solution is likely correct, but my handout gives a different solution, and I’m not very familiar with geometry. Is there anything that I’m missing?